Mathematics Term 1 STPM Chapter 2 Sequences and Series 95 2 Example 3 Determine if each of the following sequences is convergent or divergent. (a) 3, 3, 3, …, 3, … (b) 1, – 1 2 , 1 3 , – 1 4 , .... , (–1)n – 1 1 1 n 2, … (c) 1, r, r 2 , r 3 , …, rn – 1, … if (i) |r| , 1, (ii) |r| 1. Solution: (a) u = 3 lim n → ∞ un = lim n → ∞ 3 = 3. The sequence is convergent. (b) un = (–1)n – 1 1 1 n 2 lim n → ∞ un = lim n → ∞5(–1)n – 1 1 1 n 26 = (–1)n – 1 lim n → ∞1 1 n 2 = (–1)n – 1 · 0 = 0 The sequence is convergent. (c) un = rn – 1 lim n → ∞ un = lim n → ∞(rn – 1) lim n → ∞ un = 0 if |r| , 1 and lim n → ∞ un = ∞ if |r| 1. The sequence is convergent if |r| , 1 and divergent if |r| 1. Properties of the limits of sequences If lim n → ∞ un = A and lim n → ∞ vn = B, then (a) lim n → ∞(un ± vn) = lim n → ∞ un ± lim n → ∞ vn = A ± B. (b) lim n → ∞(un · vn) = lim n → ∞ un · lim n → ∞ vn = A · B. (c) lim n → ∞ 1 un vn 2 = lim n → ∞un lim n → ∞vn = A B , provided B ≠ 0. Example 4 If un and vn are the nth terms of two sequences where un = 3 and vn = 1 n respectively, find the limit of each sequence when n → ∞. Hence, find lim n → ∞ 1 3n + 1 n 2. Solution: lim n → ∞ un = lim n → ∞ 3 = 3. lim n → ∞ vn = lim n → ∞1 1 n 2 = 0
96 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 lim n → ∞ 1 3n + 1 n 2 = lim n → ∞13 + 1 n 2 = lim n → ∞3 + lim n → ∞1 1 n 2 = 3 + 0 = 3 The summation notation ∑ Suppose we want to find the sum of the series of numbers 1 + 2 + 3 + … + 100 One simpler way of representing the sum of a series such as this is by using the Greek alphabet ∑, read as “sigma”. For example, we write 100 ∑ r=1 r to represent the sum 1 + 2 + 3 + … + 100, where the r th term, ur , is r, i.e. 100 ∑ r=1 ur = 100 ∑ r=1 r = 1 + 2 + 3 + … + 100. Similarly, we write 200 ∑ r=101 to represent the sum 101 + 102 + 103 + … + 200, i.e. 200 ∑ r=101 r = 101 + 102 + 103 + … + 200. Notice that 101 + 102 + … + 200 = (100 + 1) + (100 + 2) + … + (100 + 100) So, we can also write 200 ∑ r=101 r = 100 ∑ r=1 (100 + r). In general, if ur is the rth term of a series, then n ∑ r=1 ur = u1 + u2 + u3 + … + un Notice that if k is a constant, n ∑ r=1 (kur ) = ku1 + ku2 + … + kun = k(u1 + u2 + … + un) = k n ∑ r=1 ur When u1 = u2 = u3 = … = un = k, then n ∑ r=1 k = k + k + … + k = nk n ∑ r=1 (ur + vr ) = (u1 + v1 ) + (u2 + v2 ) + … + (un + vn) = (u1 + u2 + … + un) + (v1 + v2 + … + vn) = n ∑ r=1 ur + n ∑ r=1 vr Similarly, n ∑ r=1 (ur – vr ) = n ∑ r=1 ur – n ∑ r=1 vr
Mathematics Term 1 STPM Chapter 2 Sequences and Series 97 2 Example 5 Express each of the following series by using the summation notation ∑. (a) 1 + 4 + 9 + 16 + … + 100 (b) 1 · 2 + 2 · 3 + 3 · 4 + … + 19 · 20 Solution: (a) 1 + 4 + 9 + 16 + … + 100 = 12 + 22 + 32 + 42 + … + 102 ∴ the r th term, ur = r 2 Hence, the series may be written as 10 ∑ r=1 r 2 . (b) The r th term, ur = r(r + 1). Hence, the series may be written as 19 ∑ r=1 r(r + 1). Example 6 Rewrite each of the following series by using the summation notation ∑. (a) 2 + 5 + 10 + 17 + … + 401 (b) 1 2 + 1 3 + 1 4 + … + 1 50 Solution: (a) 2 + 5 + 10 + 17 + … + 401 = (12 + 1) + (22 + 1) + (32 + 1) + (42 + 1) + … + (202 + 1) The r th term, ur = r 2 + 1, (1 r 20) Hence, the series may be written as 20 ∑ r=1 (r 2 + 1). (b) 1 2 + 1 3 + 1 4 + … + 1 50 = 1 1 + 1 + 1 2 + 1 + 1 3 + 1 + … + 1 49 + 1 The r th term, ur = 1 r + 1 , 1 r 49. Hence, the series may be written as 49 ∑ r=1 1 r + 1 . Example 7 Write down the first three terms and the last term of the following series. (a) 20 ∑ r=1 r(r + 2) (b) 10 ∑ r=1 (–1)r + 1 2r Solution: (a) The r th term, ur = r(r + 2) Thus u1 = 1(1 + 2) = 3 u2 = 2(2 + 2) = 8 u3 = 3(3 + 2) = 15 The last term, u20 = 20 (20 + 2) = 440 (b) The r th term, ur = (–1)r + 1 2r Thus u1 = (–1)2 21 = 2 u2 = (–1)3 22 = –4 u3 = (–1)4 23 = 8 The last term, u10 = (–1)11 210 = –1024
98 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 Exercise 2.1 1. In each of the sequences below, write down the next two terms and state the nth term. (a) 5, 10, 15, 20, … (b) 4, 7, 10, 13, … (c) 2, 5, 8, 11, … (d) 2, 4, 8, 16, … (e) 3, – 3 5 , 3 25 , – 3 125 , … (f) 1 3 , 3 5 , 5 7 , 7 9 , … (g) 1, 5 2 , 25 4 , 125 8 , … (h) 1 2 , 1 6 , 1 12 , 1 20 , … (i) 1, –2, 3, –4, … 2. Write down the first four terms of each of the following sequences. (a) 5 n n + 16 (b) 5 (–1)n + 1 n! 6 (c) 5 (2x) n – 1 (2n – 1)2 6 (d) 5 (–1)n x2n – 1 (2n – 1)(2n + 1) 6 (e) 5 cos nx nxn 6 3. Find the nth term of each of the following sequences. (a) – 1 5 , 3 8 , – 5 11 , 7 14 , – 9 17 , … (b) 1, 0, 1, 0, 1, … (c) 2 3 , 0, 3 4 , 0, 4 5 , … 4. Find the next five terms of each of the sequences below, defined by the recursive formula given. (a) un + 1 = nun – n, u1 = 3. (b) un + 1 = 2un – (–1)n , u1 = 1. (c) un + 2 = 3un + 1 – 2un, u1 = 1, u2 = 3. 5. Determine if each of the following sequences are convergent or divergent. For those that are convergent, find their limits. (a) 1 + 1 2 , 1 + 1 3 , 1 + 1 4 , … (b) 1 + 1 2 , 2 + 1 3 , 3 + 1 4 , … 2 2 2 (c) , , , … (d) 3 – 1 2n , 3 – 2 3n , 3 – 3 4n , … 1 + 1 2 1 + 1 3 1 + 1 4 6. Evaluate (a) lim n → ∞1 3 n 2 (b) lim n → ∞1 2 n – 12 (c) lim n → ∞1 n2 + 2 n 2 (d) lim n → ∞1 n + 2 n – 12 (e) lim n → ∞1 2n2 + 2n 5n2 – n 2 (f) lim n → ∞1 1 + 2 × 5n 4 + 3 × 5n 2 7. Write each of the following series by using the ∑ notation. (a) 1 + 8 + 27 + 64 + … + 1 000 (b) 1 + 3 + 5 + 7 + … + 99 (c) 1 + 1 2 + 1 4 + 1 8 + … + 1 512 (d) 1 – 1 3 + 1 9 – 1 27 + 1 81 – 1 243 (e) 11 + 8 + 5 + 2 – 1 – 4 – 7 (f) 5 + 7 + 11 + 19 + 35 + 67 (g) –x + 2x2 – 3x3 + … + 10x10 (h) 1 2 · 3 + 2 3 · 4 + 3 4 · 5 + … + 8 9 · 10 8. Write down the terms of each of the following series without the ∑ notation. (a) 10 ∑ r=1 (r 2 – r) (b) 8 ∑ r=1 1 r 2 (c) 5 ∑ r=1 (3r + 2) (d) 9 ∑ r=1 (–1)r – 1(3r) (e) 10 ∑ r=1 (8 – r) (f) 7 ∑ r=1 (r + 2)(r + 4) (g) 5 ∑ r=1 (6 – r) 2 (h) 5 ∑ r=1 (2r 2 – 3r – 5)
Mathematics Term 1 STPM Chapter 2 Sequences and Series 99 2 2.2 Series Arithmetic series Consider the sequence of numbers 2, 5, 8, 11, …, 29 Each term (except the first term) in this sequence can be obtained by adding a fixed number 3 to the term before it. Hence, the above sequence can also be written as 2, [2 + 1(3)], [2 + 2(3)], [2 + 3(3)], …, [2 + 9(3)]. A sequence such as this is called an arithmetic progression and the fixed number is called the common difference. For the arithmetic progression 10, 7, 4, 1, –2, …, –17 the common difference is 7 – 10, i.e. –3. Hence, the arithmetic progression can also be written as 10, [10 + 1(–3)], [10 + 2(–3)], [10 + 3(–3)], …, [10 + 9(–3)]. If the first term of an arithmetic progression is a and the common difference is d, then this arithmetic progression can be represented by a, (a + d), (a + 2d), (a + 3d), …, [a + (n – 1)d]. and the nth term is un = a + (n – 1)d Example 8 Given that the fifth term of an arithmetic progression is 21 and its tenth term is 41, find the common difference, first term and the nth term of this arithmetic progression. Solution: Let a be the 1st term and d the common difference. So, the 5th term is a + 4d, i.e. a + 4d = 21 ………… The 10th term is a + 9d, i.e. a + 9d = 41 ………… – : 5d = 20 d = 4 Substituting d = 4 into : a + 16 = 21 a = 5 Hence, the 1st term is 5 and the common difference is 4. The nth term is un = a + (n – 1)d = 5 + (n – 1)4 = 1 + 4n
100 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 Example 9 The nth term of an arithmetic progression is 1 10 (5n – 3). Obtain the first two terms and also the common difference of the arithmetic progression. Solution: The nth term, un = 1 10 (5n – 3) The 1st term, u1 = 1 10 (5 – 3) = 1 5 The 2nd term, u2 = 1 10 (5 × 2 – 3) = 7 10 The common difference, d = u2 – u1 = 7 10 – 1 5 = 1 2 Sum of a finite arithmetic series When the terms of an arithmetic progression are added up, we will obtain an arithmetic series. Consider the following arithmetic series made up of 10 terms, i.e. S10 = 3 + 6 + 9 + 12 + … + 27 + 30 ………… By rewriting the terms of the series backwards, we obtain S10 = 30 + 27 + 24 + 21 + … + 6 + 3 ………… Adding, +: 2S10 = 33 + 33 + 33 + 33 + … + 33 + 33 = 33 × 10 (since the series has 10 terms) = 330 ∴ S10 = 165 For any arithmetic series with first term a, common difference d and nth term l, the series can be written as Sn = a + (a + d) + (a + 2d) + … + (l – d) + l ………… By rewriting the terms of the series backwards, we obtain Sn = l + (l – d) + (l – 2d) + … + (a + d) + a ………… + : 2Sn = (a + l) + (a + l) + (a + l) + … + (a + l) + (a + l) = n(a + l) since the series has n terms ∴ n Sn = — (a + l). 2 Since the nth term, l = a + (n – 1) d, n Sn = — [a + a + (n – 1) d] 2 n i.e. Sn = — [2a + (n – 1) d] 2 Both these formulae derived can be used to find the sum of the first n terms of an arithmetic series. Arithmetic Sequence VIDEO
Mathematics Term 1 STPM Chapter 2 Sequences and Series 101 2 Example 10 Find the sum of the integers which can be divided by 6 exactly and which lie in between 50 and 150. Solution: The 1st term divisible by 6 is 54. The last term divisible by 6 is 144. The required series is Sn = 54 + 60 + 66 + … + 144 This is an arithmetic progression with a = 54 and d = 60 – 54 = 6. The last term, l = a + (n – 1)d = 144 Hence, 54 + (n – 1)6 = 144 6n = 96 n = 16 Hence, the sum of this series is S16 = 16 2 (a + l) = 16 2 (54 + 144) = 8 × 198 = 1584 Example 11 In an arithmetic progression, the sum of the first ten terms is 520 and the 7th term is twice the 3rd term. Find the first term, a, and the common difference, d. Solution: The sum of the first ten terms is S10 = 10 2 (2a + 9d) = 520 2a + 9d = 104 ………… The 7th term is a + 6d and the 3rd term is a + 2d. a + 6d = 2(a + 2d) a = 2d ………… Substituting into : 4d + 9d = 104 13d = 104 d = 8 a = 16 Hence, for the given arithmetic progression, the 1st term is 16 and the common difference is 8. Example 12 Find the sum of the first 10 terms of the series n ∑ r=1 ln 3r , and the smallest value of n such that the sum of the first n terms exceeds 1000. Solution: By substituting r = 1, 2, 3, … into ln 3r , we have n ∑ r=1 ln 3r = ln 3 + ln 32 + ln 33 + … + ln 3n = ln 3 + 2 ln 3 + 3 ln 3 + … + n ln 3.
102 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 This series is an arithmetic series with 1st term, a = ln 3 and common difference d = ln 3, i.e. a = d = ln 3. 10 ∑ r=1 ln 3r = ln 3 + 2 ln 3 + 3 ln 3 + … + 10 ln 3 = (1 + 2 + 3 + … + 10) ln 3 = 10 2 (1 + 10) ln 3 = 55 ln 3 and n ∑ r=1 ln 3r = (1 + 2 + 3 + … + n) ln 3 = n 2 (1 + n) ln 3. We have to find the smallest value of n such that n 2 (1 + n) ln 3 1000 n(1 + n) 2000 ln 3 n(1 + n) 1821 By using the calculator, we find that if n = 42, n(1 + n) = 42 × 43 = 1806 , 1821 if n = 43, n(1 + n) = 43 × 44 = 1892 1821 Hence, the smallest value of the integer n such that n ∑ r=1 ln 3r 1000 is 43. Example 13 The nth term of an arithmetic progression is 32 and its first term is 2. If the sum of the first n terms is 357, find the value of n. If the smallest term in the series which exceeds 100 is the kth term, find the value of k. Solution: The 1st term, a = 2 The nth term, a + (n – 1)d = 32 d = common difference 2 + (n – 1)d = 32 (n – 1)d = 30 ………… Sum of the 1st n terms is 357. Thus n 2 [2a + (n – 1)d] = 357 i.e. n 2 (4 + 30) = 357 17n = 357 n = 21 Substitute n = 21 into , 20d = 30 d = 3 2 The kth term, uk = a + (k – 1)d = 2 + (k – 1) · 3 2 = 1 2 (3k + 1) For 1 2 (3k + 1) 100 3k + 1 200 3k 199 k 66.3 Hence, the smallest term in the series which exceeds 100 is the 67th term, i.e. k = 67.
Mathematics Term 1 STPM Chapter 2 Sequences and Series 103 2 Example 14 The sum of the first n terms of an arithmetic progression is given by Sn = pn + qn2 . Given that S3 = 6 and S5 = 11, (a) find the values of p and q, (b) deduce, or otherwise, an expression for the nth term and the value of its common difference. Solution: (a) Sn = pn + qn2 When n = 3, S3 = 3p + 9q = 6 p + 3q = 2 ………… When n = 5, S5 = 5p + 25q = 11 p + 5q = 11 5 ………… – : 2q = 1 5 q = 1 10 Substituting q = 1 10 into , p + 3 10 = 2 p = 2 – 3 10 = 17 10 Hence, p = 17 10 and q = 1 10 . (b) Sn = pn + qn2 Thus Sn – 1 = p(n – 1) + q(n – 1)2 The nth term, un = Sn – Sn – 1 = pn + qn2 – p(n –1) – q(n – 1)2 = pn + qn2 – pn + p – q(n2 – 2n + 1) = p + 2nq – q = 17 10 + 2n1 1 102 – 1 10 = 1 5 (n + 8) The common difference, d = un – un – 1 = 1 5 (n + 8) – 1 5 (n – 1 + 8) = 1 5 (n + 8) – 1 5 (n + 7) = 1 5 Check: u2 = 1 5 (2 + 8) = 10 5 u1 = 1 5 (1 + 8) = 9 5 d = u2 – u1 = 10 5 – 9 5 = 1 5
104 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 Arithmetic mean If a, b and c are three consecutive terms of an arithmetic progression, the term b is called the arithmetic mean of a and c. The common difference = b – a = c – b ∴ 2b = a + c b = 1 2 (a + c) The arithmetic mean of two numbers, p and q, is 1 2 (p + q). Exercise 2.2 1. For each of the following arithmetic series, write down the term indicated in brackets, and the nth term. (a) 7 + 11 + 15 + … (7th term) (b) 18 + 11 + 4 + … (6th term) (c) –7 – 5 – 3 – … (20th term) (d) 3 + 3 2 3 + 4 1 3 + … (15th term) 2. Find the sum of the terms of each of the following series. (a) 5 + 9 + 13 + … + 81 (b) 85 + 82 + 79 + … + 13 (c) –22 – 17 – 12 – … + 68 (d) 1 + 1 3 5 + 2 1 5 + … + 18 2 5 3. Find the sum of each of the following arithmetic series. (a) 3 + 8 + 13 + … to 18th term (b) 4 + 7 1 2 + 11 + … to 20th term (c) 21 + 18 + 15 + … to 12th term (d) –15 – 9 – 3 – … to 10th term 4. Find the sum of the arithmetic series –11 – 7 – 3 + 1 + … from the 11th term to the 20th term. 5. Find the sum of the odd numbers between 0 and 500 which are divisible by 7. 6. Find the sum of the numbers between 1 and 200 inclusive, which are not divisible by 6. 7. The first and last terms of an arithmetic series are 29 and 179 respectively. If the total number of terms is 25, find the common difference and the sum of the series. 8. The second and seventh terms of an arithmetic series are –5 and 10 respectively. Find the eighth term and the smallest value of n such that the sum of n terms exceeds 500. 9. In an arithmetic series, the sum of the first 15 terms is 615, and the 13th term is 6 times the 2nd term. Find the first three terms. 10. The sum of the first n terms of a series is 3n2 + n. Show that the series is an arithmetic progression, and find the first term and common difference. 11. In an arithmetic progression, the sum of the first 2n terms is equal to the sum of the following n terms. If the first term is 12 and common difference is 3, find the value of n. 12. Show that the sum of the odd integers from 1 till (2n – 1) is n2 . Find the smallest value of n such that the sum of n terms exceeds 4 000. 13. Find the arithmetic mean of (a) 3 and 27 (b) 3 and –27 (c) 1 3 and 1 27 (d) log10 3 and log10 27
Mathematics Term 1 STPM Chapter 2 Sequences and Series 105 2 Geometric series Consider the sequence of numbers 1, 2, 4, 8, 16, … Each term (except the first term) in the sequence is obtained by multiplying the previous term with a fixed number 2. Thus, this sequence can also be written as 1, 1 × 2, 1 × 22 , 1 × 23 , 1 × 24 , … This type of sequence is called a geometric progression and the fixed number is called common ratio. If a geometric progression has first term 3 and common ratio –2, then the terms are 3, 3(–2), 3(–2)2 , 3(–2)3 , … or 3, –6, 12, –24, … If the first term of a geometric progression is a and its common ratio is r, then the geometric progression may be represented as a, ar, ar2 , …, ar n – 1 with its nth term, un = ar n – 1 Sum of a finite geometric series When the terms of a geometric progression are added up, we will obtain a geometric series. Consider the following geometric series which is made up of 10 terms, with the first term 1 and common ratio 5, i.e. S10 = 1 + 1(5) + 1(5)2 + 1(5)3 + … + 1(5)9 S10 = 1 + 5 + 52 + 53 + … + 59 ………… × 5: 5S10 = 5 + 52 + 53 + 54 + … + 510 ………… – : (5 – 1) S10 = 510 – 1 S10 = 510 – 1 5 – 1 = 1 4 (510 – 1) For any geometric series with first term a and common ratio r ≠ 1, the sum of the first n terms, Sn, can be written as Sn = a + ar + ar2 + ar3 + … + arn – 1……… × r: rSn = ar + ar2 + ar3 + ar4 + … + arn ……… – : (r – 1)Sn = arn – a or Sn = a(rn – 1) r – 1 , for r . 1 Sn = a(1 – rn ) 1 – r , for r , 1 Geometric Sequence VIDEO
106 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 Example 15 The 5th term of a geometric series is 18 and the 7th term is 54. Given that the sum of the first 10 terms of this series is positive, find the 1st term, the common ratio and the sum of the first 10 terms. Solution: Let a be the 1st term and r the common ratio. the 5th term, ar4 = 18 ………… and the 7th term, ar6 = 54 ………… ÷ : r 2 = 54 18 = 3 r = ± 3 Substituting r = ± 3 into , a(9) = 18 a = 2 From the formula Sn = a(rn – 1) r – 1 , when r = 3 , S10 = 2[( 3)10 – 1] 3 – 1 = 2(243 – 1) 3 – 1 = 484 3 – 1 When r = – 3 , S10 = 2[1 – (– 3)10] 1 – (– 3 ) = 2(1 – 243) 1 + 3 = – 484 3 + 1 Since it is given that S10 0, thus a = 2, r = 3 and S10 = 484 3 – 1 . Example 16 The first term of a geometric series is 5 and the common ratio is 1.5. Find the number of terms needed such that the sum of the series exceeds 200. Solution: Given that a = 5 and r = 1.5 The sum of the 1st n terms is Sn = 5[(1.5)n – 1] 1.5 – 1 = 10[(1.5)n – 1] For Sn to exceed 200, 10[(1.5)n – 1] 200 (1.5)n – 1 20 (1.5)n 21 n log10 1.5 log10 21 n log10 21 log10 1.5 = 7.5 Hence, the smallest value of n is 8, i.e. 8 terms are needed such that its sum exceeds 200.
Mathematics Term 1 STPM Chapter 2 Sequences and Series 107 2 Geometric mean If a, b and c are three consecutive terms of a geometric progression, the term b is the geometric mean of a and c. The common ratio = —b c = — a b ∴ b2 = ac b = (ac) The geometric mean of two numbers, p and q, is (pq). Example 17 If (x + 1), 2 2 and (3x – 2) are three consecutive terms of a geometric progression, find the integral value of x. Solution: (x + 1), 2 2 and (3x – 2) forms a geometric progression. Thus, 2 2 is the geometric mean of (x + 1) and (3x – 2) i.e. 2 2 = (x + 1)(3x – 2) 8 = (x + 1)(3x – 2) 8 = 3x2 + x – 2 3x2 + x – 10 = 0 (3x – 5)(x + 2) = 0 x = 5 3 or –2 Hence, the integral value of x is –2. Exercise 2.3 1. For each of the following geometric series, write down the term indicated in brackets and the nth term. (a) 1 2 + 1 + 2 + … (8th term) (b) 162 + 54 + 18 + … (6th term) (c) 200 – 50 + 12 1 2 + … (5th term) (d) – 4 9 – 2 3 – 1 – … (7th term) 2. Find the number of terms in each of the geometric series below, and also the sum of the series. (a) 1 4 + 1 2 + … + 64 (b) 9 – 6 + … + 256 729 (c) 100 + 50 + … + 25 16 (d) 2 – 8 3 + … – 14 2150 2187 3. Find the sum of each of the following geometric series. (a) 100 + 20 + … to 8th term (b) 4 – 2 + … to 10th term (c) 2 – 6 + … to nth term (d) ak + ak + 2 + … to nth term 4. The terms of a geometric series are positive, with the first term 80. If the sum of the first three terms is 185, find the common ratio of the series. 5. If two numbers, m and n, are such that m, n and 10 form an arithmetic progression, whereas n, m and 10 form a geometric progression, find the values of m and n.
108 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 6. Find the geometric mean of each of the following numbers. (a) 3 and 27 (b) 1 3 and 1 27 (c) 103 and 1027 7. Given that the geometric mean of 4p – 3 and 9p + 4 is 6p – 1, find the values of p. 8. The second and fifth terms of a geometric series are 405 and –120 respectively. Find the seventh term and the sum of the first seven terms of the series. 9. In a geometric progression, the second term exceeds the first term by 20 and the fourth term exceeds the second term by 15. Find the possible values of the first term. 10. Find the sum of the first n terms of the series 1 12 + 1 4 + 3 4 + …. Find also the number of terms required such that the sum exceeds 100. 11. A geometric series has first term 16 and common ratio 3 4 . If the sum of the first n terms exceeds 60, find the smallest value of n. Sum of an infinite geometric series When the number of terms of a geometric series is infinite, it is called an infinite geometric series. Consider the infinite geometric series 1 + 1 2 + 1 4 + … + 1 1 2 2 n – 1 + … The sum of the first n terms is 131 – 1 1 2 2 n 4 Sn = 1 – 1 2 = 2 31 – 1 1 2 2 n 4 As n increases, 1 1 2 2 n → 0 and Sn → 2. We say the limit of Sn as n → ∞ exists and equals 2, and we write lim n → ∞Sn = 2. This series is called a convergent series with a sum of 2. Notice that the common ratio is 1 2 , 1. We now consider the infinite geometric series 1 + 2 + 4 + … + 2n + … The sum of this series is infinite. This series is called a divergent series. In general, for a geometric series with first term a and common ratio r, Sn = a(1 – r n ) 1 – r = a 1 – r – 1 a 1 – r2rn
Mathematics Term 1 STPM Chapter 2 Sequences and Series 109 2 S∞ = lim n → ∞ Sn = lim n → ∞ 3 a 1 – r – 1 a 1 – r2rn 4 = a 1 – r – 1 a 1 – r2 lim n → ∞ rn If |r| 1, lim n → ∞ rn → ∞, and the series is divergent. If |r| , 1, lim n → ∞ rn → 0, and the series is convergent with sum to infinity S∞ = a 1 – r . When |r| , 1, the geometric series 1 + a + ar + ar2 + … is convergent, with the sum to infinity S∞ = a 1 – r . Example 18 Find the sum of each of the following series. (a) 18 – 6 + 2 – … (b) 1 – 1 4 + 1 16 – … Solution: (a) 18 – 6 + 2 + … = 18 + 181– 1 3 2 + 181– 1 3 2 2 + … This is a geometric series with 1st term a = 18. Common ratio r = – 1 3 |r| = u– 1 3 u = 1 3 , 1 Sum to infinity S∞ = a 1 – r 18 = 1 – 1– 1 3 2 = 54 4 = 13 1 2 (b) 1 – 1 4 + 1 16 – … = 1 + 11– 1 4 2 + 11– 1 4 2 2 + … This is a geometric series with 1st term a = 1. Common ratio r = – 1 4 |r| = |– 1 4 | = 1 4 , 1 Sum to infinity S∞ = a 1 – r 1 = 1 – 1– 1 4 2 = 4 5
110 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 Example 19 Express the recurring decimal 0.327 as an infinite geometric series. Hence, express 0.327 as a fraction in its simplest form. (Note: The recurring decimal 0.327 = 0.327327327…) Solution: 0.327 = 0.327327327… = 0.327 + 0.000327 + 0.000000327 + … = 0.327 + 0.327 × 10–3 + 0.327 × 10–6 + … This is an infinite geometric series with a = 0.327 and r = 10–3 = 0.001. Hence, 0.327 = 0.327 1 – 0.001 a Using S∞ = ––––– 1 – r = 0.327 0.999 = 327 999 = 109 333 Example 20 Find the sum of the series 1 + 3x + 9x2 + 27x3 + …, stating the range of values of x for which the result is valid. Solution: 1 + 3x + 9x2 + 27x3 + … is a geometric series with a = 1 and r = 3x. Hence, the sum to infinity is S∞ = 1 1 – 3x . The result is valid if |3x|, 1, i.e. |x| , 1 3 or – 1 3 , x , 1 3 Example 21 The first term of a geometric series is 2 and the common ratio is 0.95. The sum of the first n terms of this series is Sn and the sum of this series is S∞. Find the smallest value of n such that S∞ – Sn , 1. Solution: The first term, a = 2 Common ratio, r = 0.95 Using the formulas Sn = a(1 – r n ) 1 – r and S∞ = a 1 – r , S∞ – Sn , 1 ⇒ a 1 – r – a(1 – r n ) 1 – r , 1 a 1 – r – a 1 – r + arn 1 – r , 1 arn 1 – r , 1
Mathematics Term 1 STPM Chapter 2 Sequences and Series 111 2 Substituting the values a = 2 and r = 0.95. 2(0.95)n 1 – 0.95 , 1 (0.95)n , 0.025 n log10 (0.95) , log10 (0.025) n(–0.0223) , –1.6021 n –1.6021 –0.0223 = 71.8 i.e. the smallest value of n such that S∞ – Sn , 1 is 72. Exercise 2.4 1. Determine whether each of the following series is convergent or otherwise. (a) 2 + 2 3 + 2 32 + … (b) 18 + 15 + 12 + … (c) 40 – 20 + 10 – … (d) 4 5 + 8 15 + 16 45 + … (e) k + 2k + 3k + … (f) 7 – 3 1 2 + 1 3 4 – … 2. Find the sum of each of the following geometric series. (a) 6 + 2 + 2 3 + … (b) 1 – 1 2 + 1 4 – … (c) 10 + 1 + 0.1 + … (d) 45 – 30 + 20 – … 3. Express each of the following recurring decimals as an infinite geometric series or as the sum of a constant and an infinite geometric series. Hence, express each of the decimals as a fraction in its simplest form. · · · · · · · · (a) 0.48 (b) 0.072 (c) 0.5813 · · · · · · (d) 0.3354 (e) 0.9218 4. Find the sum of each of the following infinite geometric series and state the range of values of x for which the result is valid. (a) 1 + 2x + 4x2 + … (b) 1 – 1 2 x + 1 4 x2 – … (c) 3 – 6x + 12x2 – … (d) x + 1 3 x2 + 1 9 x3 + … 5. The sum of the first n terms of a geometric series is 2 3n (3n – 1). Obtain the first three terms and the sum to infinity of the series. 6. Find the value of r such that the sum of the series 1 + r + r 2 + … + rn – 1 + … is twice the sum of the series 1 – r + r 2 – r 3 + … 7. A ball rebounds to a height five-eighths of its previous height above the ground. If a ball is dropped from a height of 3 metres, find the total distance travelled by the ball before it comes to rest. 8. The first three terms of a geometric series are 2, – 1 2 and 1 8 respectively. Find the sum to infinity of the series. Find the smallest value of n such that the difference between the sum of the first n terms and the sum to infinity is less than 10–5.
112 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 9. Find the sum to the nth term of the geometric series 108 + 60 + 33 1 3 + …. If s is the smallest number which exceeds this sum for all values of n, find the value of s. Find also the smallest value of n such that the sum of the series exceeds 99% of the value of s. 10. Write down the first four terms of a geometric series with sum 4 3 and first term 1. Find the value of r such that the series r 2 + r 2 1 + r 2 + r 2 (1 + r 2 ) 2 + … is convergent, and also the non zero value of the sum. Summation of finite series involving powers of integers Consider the series of the first n positive integers 1 + 2 + 3 + … + (n – 1)+ n. This series can also be written as n ∑ r=1 r = 1 + 2 + 3 + … + (n – 1) + n ………… By rewriting this series backwards term by term, n ∑ r=1 r = n + (n – 1) + (n – 2) + … + 2 + 1 ………… + : 2 n ∑ r=1 r = (n + 1) + (n + 1) + (n + 1) + … + (n + 1) + (n + 1) = n(n + 1) Since the series has n terms n ∑ r=1 r = 1 2 n(n + 1) Now, consider the series of the squares of the first n positive integers, i.e. n ∑ r=1 r 2 = 12 + 22 + 32 + … + (n – 1)2 + n2 . By using the identity (r + 1)3 r3 + 3r 2 + 3r + 1, we have (r + 1)3 – r3 = 3r 2 + 3r + 1. By adding the terms of the identity one by one with values of r from r = n to r = 1 , we get n ∑ r=1 [(r + 1)3 – r 3 ] = 3 n ∑ r=1 r 2 + 3 n ∑ r=1 r + n ∑ r=1 1 The LHS of this identity can be written as [(n + 1)3 – n3 ] + [n3 – (n – 1)3 ] + … + (33 – 23 ) + (23 – 13 ) = (n + 1)3 – 13 = (n + 1)3 – 1 Thus, (n + 1)3 – 1 = 3 n ∑ r=1 r 2 + 3 n ∑ r=1 r + n ∑ r=1 1 3 n ∑ r=1 r 2 = (n + 1)3 – 1 – 3 n ∑ r=1 r – n ∑ r=1 1 = (n + 1)3 – 1 – 3 2 n(n + 1) – n = (n + 1)3 – (n + 1) – 3 2 n(n + 1)
Mathematics Term 1 STPM Chapter 2 Sequences and Series 113 2 = 1 2 (n + 1)[2(n + 1)2 – 2 – 3n] = 1 2 (n + 1)(2n2 + 4n + 2 – 2 – 3n) = 1 2 (n + 1)(2n2 + n) = 1 2 n(n + 1)(2n + 1) Thus n ∑ r=1 r 2 = 1 6 n(n + 1)(2n + 1) Similarly, by using the identity (r + 1)4 – r 4 = 4r 3 + 6r 2 + 4r + 1, we get n ∑ r=1 r 3 = 1 4 n2 (n + 1)2 = 3 1 2 n(n + 1)4 2 = 1 n ∑ r=1 r2 2 Sum of the first n positive integers is n ∑ r=1 r = 1 2 n(n + 1). Sum of the squares of the first n positive integers is n ∑ r=1 r 2 = 1 6 n(n + 1)(2n + 1) Sum of the cubes of the first n positive integers is n ∑ r=1 r 3 = 1 4 n2 (n + 1)2 . Example 22 Evaluate (a) 100 ∑ r=1 r (b) 50 ∑ r=1 r 2 (c) 25 ∑ r=1 r 3 Solution: (a) By using n ∑ r=1 r = 1 2 n(n + 1) and substituting n = 100, 100 ∑ r=1 r = 1 2 (100)(101) = 5050 (b) By usingn ∑ r=1 r 2 = 1 6 n(n + 1)(2n + 1) and substituting n = 50, 50 ∑ r=1 r 2 = 1 6 (50)(51)(101) = 42 925
114 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 (c) By using n ∑ r=1 r 3 = 1 4 n2 (n + 1)2 and substituting n = 25, 25 ∑ r=1 r 3 = 1 2 (25)2 (26)2 = 105 625 Example 23 Find the sum of the series 12 + 32 + 52 + 72 + … to n terms. Hence, find the value of 12 + 32 + 52 + … + 992 . Deduce the value of 512 + 532 + … + 992 . Solution: The sum of the squares of the odd integers is 12 + 32 + 52 + 72 + … The r th term is (2r – 1)2 Hence, the sum of the series is n ∑ r=1 (2r – 1)2 . n ∑ r=1 (2r – 1)2 = n ∑ r=1 (4r 2 – 4r + 1) = 4 n ∑ r=1 r 2 – 4 n ∑ r=1 r + n ∑ r=1 1 = 43 n 6 (n + 1)(2n + 1)4 – 43 n 2 (n + 1)4 + n = n 3 [2(n + 1)(2n + 1) – 6(n + 1) + 3] = n 3 (4n2 + 6n + 2 – 6n – 6 + 3) = n 3 (4n2 – 1) 12 + 32 + 52 + … + 992 = 50 ∑ r=1 (2r – 1)2 nth term, 2n – 1 = 99 n = 50 = 50 3 [4(50)2 – 1] = 50 3 × 9999 = 166 650 512 + 532 + … + 992 = (12 + 32 + … + 992 ) – (12 + 32 + … + 492 ) = 50 ∑ r=1 (2r – 1)2 – 25 ∑ r=1 (2r – 1)2 = 166 650 – 25 3 [4(25)2 – 1] = 166 650 – 20 825 = 145 825
Mathematics Term 1 STPM Chapter 2 Sequences and Series 115 2 Example 24 Find the r th term of the series 1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + … By using the results for n ∑ r=1 r and n ∑ r=1 r 2 , find the sum of the first n terms of the above series. Solution: The series is 1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + … The r th term is r(r + 1)(r + 2). The sum of the 1st n terms of the series is n ∑ r=1 r(r + 1)(r + 2) = n ∑ r=1 (r 3 + 3r 2 + 2r) = n ∑ r=1 r 3 + 3 n ∑ r=1 r 2 + 2 n ∑ r=1 r = n2 4 (n + 1)2 + 33 n 6 (n + 1)(2n + 1)4 + 23 n 2 (n + 1)4 = n 4 (n + 1)[n(n + 1) + 2(2n + 1) + 4] = n 4 (n + 1)(n2 + 5n + 6) = n 4 (n + 1)(n + 2)(n + 3) Summation of series using the method of differences Consider the sum of the terms of a series such as u1 + u2 + u3 + … + un Suppose that the r th term, ur , can be expressed in the form f(r) – f(r – 1), where f(r) is a function of r. Then n ∑ r=1 ur = n ∑ r=1 [f(r) – f(r – 1)] = f(n) – f(n – 1) + f(n – 1) – f(n – 2) + f(n – 2) – f(n – 3) + … + f(3) – f(2) + f(2) – f(1) + f(1) – f(0) = f(n) – f(0) This means that if the r th term, ur , of a series can be expressed as the difference of a function of r and a function of (r – 1), then the sum of the series, n ∑ r=1 ur , can be determined. This method of finding the sum of a series is called the method of differences. If ur = f(r) – f(r – 1), then n ∑ r=1 ur = f(n) – f(0) Substituting the values of r, term by term, with r = n, n – 1, n – 2, …, 3, 2, 1
116 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 Example 25 Show that r(r + 1)(r + 2)(r +3) – (r – 1)r(r + 1)(r + 2) = 4r(r + 1)(r + 2). Hence, find n ∑ r=1 r(r + 1)(r + 2). Solution: r(r + 1)(r + 2)(r + 3) – (r – 1)r(r + 1)(r + 2) = r(r + 1)(r + 2)[(r + 3) – (r – 1)] = 4r(r + 1)(r + 2) Let f(r) = r(r + 1)(r + 2)(r + 3) Then f(r – 1) = (r – 1)r(r + 1)(r + 2) f(r) – f(r – 1) = r(r + 1)(r + 2)(r + 3) – (r – 1)r(r + 1)(r + 2) = 4r(r + 1)(r + 2) For the series n ∑ r=1 r(r + 1)(r + 2), ur = r(r + 1)(r + 2). f(r) – f(r – 1) = 4ur or ur = 1 4 [f(r) – f(r – 1)] ∴ n ∑ r=1 ur = 1 4 n ∑ r=1 [f(r) – f(r – 1)] = 1 4 [f(n) – f(0)] = 1 4 [n(n + 1)(n + 2)(n + 3) – 0] Thus, n ∑ r=1 r(r + 1)(r + 2) = 1 4 n(n + 1)(n + 2)(n + 3). Note: Compare the method of differences shown in Example 25 with the method shown in Example 24. The method of differences can also be used if an expression can be expressed in partial fractions, as shown in the following example. Example 26 Express 1 r(r + 1) in partial fractions. Hence, find n ∑ r=1 1 r(r + 1) . Solution: Let 1 r(r + 1) ≡ A r + B r + 1 ∴ 1 ≡ A(r + 1) + Br Let r = 0 : 1 = A Let r = –1 : 1 = –B B = –1 Thus 1 r(r + 1) = 1 r – 1 r + 1
Mathematics Term 1 STPM Chapter 2 Sequences and Series 117 2 n ∑ r=1 1 r(r + 1) = n ∑ r=1 1 1 r – 1 r + 1 2 = – n ∑ r=1 1 1 r + 1 – 1 r 2 Let f(r) = 1 r + 1 , f(r – 1) = 1 r Thus n ∑ r=1 1 r(r + 1) = – n ∑ r=1 [f(r) – f(r – 1)] = –[f(n) – f(0)] = –1 1 n + 1 – 12 = 1 – 1 n + 1 = n n + 1 Example 27 Show that 1 (r + 1)(r + 2) – 1 r(r + 1) = – 2 r(r + 1)(r + 2) . Using the method of differences, find n ∑ r=1 1 r(r + 1)(r + 2) . Solution: 1 (r + 1)(r + 2) – 1 r(r + 1) = r – (r + 2) r(r + 1)(r + 2) = – 2 r(r + 1)(r + 2) Thus 1 r(r + 1)(r + 2) = – 1 2 3 1 (r + 1)(r + 2) – 1 r(r + 1) 4 Let ur = 1 r(r + 1)(r + 2) and f(r) = 1 (r + 1)(r + 2) . ∴ ur = – 1 2 [f(r) – f(r – 1)]. and n ∑ r=1 ur = – 1 2 n ∑ r=1 [f(r) – f(r – 1)] = – 1 2 [f(n) – f(0)] = – 1 2 3 1 (n + 1)(n + 2) – 1 2 4 n ∑ r=1 1 r(r + 1)(r + 2) = 1 2 3 1 2 – 1 (n + 1)(n + 2)4.
118 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 Example 28 Find the r th term, ur , of the series 2 · 1! + 5 · 2! + 10 · 3! + 17 · 4! + … + (n2 + 1)·n! By expressing ur as the difference of two functions of r, find the sum of the above series. Solution: For the series 2 · 1! + 5 · 2! + 10 · 3! + 17 · 4! + … + (n2 + 1) · n!, the r th term, ur = (r 2 + 1) · r! = [r(r + 1) – (r – 1)] · r! = r(r + 1)r! – (r – 1) · r! = r(r + 1)! – (r – 1)r! (r + 1)r! = (r + 1)! Let f(r) = r(r + 1)! and f(r – 1) = (r – 1)r! Then ur = f(r) – f(r – 1) and n ∑ r=1 ur = n ∑ r=1 [f(r) – f(r – 1)] = f(n) – f(0) = n(n + 1)! – 0 ∴ n ∑ r=1 (r 2 + 1) r! = n(n + 1)! Exercise 2.5 1. Evaluate (a) 4 ∑ r=1 (r 3 + 3r) (b) 12 ∑ r=10 (100 – r 2 ) (c) 8 ∑ r=3 (4r + 5) (d) 5 ∑ r=1 90 r (e) 8 ∑ r=1 sin rπ 3 (f) 5 ∑ r=0 (–1)r (1 + 2r + 1) 2. Using the results for n ∑ r=1 r, n ∑ r=1 r 2 and n ∑ r=1 r 3 , find the value of n ∑ r=1 ur for each of the following cases. (a) ur = 2r 3 – r + 1 (b) ur = r 2 (r + 2) (c) ur = (r + 2)(r + 3)(2r – 1) (d) ur = (r + 1)(r + 3)(r + 5) 3. For each of the following series, write down its r th term. Hence, find the sum up to the nth term. (a) 1 · 4 + 2 · 7 + 3 · 10 + … (b) 12 · 5 + 22 · 6 + 32 · 7 + … (c) 1 · 3 + 2 · 4 + 3 · 5 + … (d) 1 · 4 · 7 + 4 · 7 · 10 + 7 · 10 · 13 + … (e) 1 1 · 2 · 3 + 1 2 · 3 · 4 + 1 3 · 4 · 5 + … (f) 1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + … 4. Given that 12 + 22 + 32 + … + n2 = 1 6 n(n + 1)(2n + 1), find the sum of the series 22 + 42 + 62 + … + 502 Using the above result, find the value of 12 + 32 + 52 + … + 492 . 5. Given that 13 + 23 + 33 + … + n3 = 1 4 n2 (n + 1)2 , find the sum of the first twenty terms of the series 2 + 16 + 54 + 128 + 250 + …
Mathematics Term 1 STPM Chapter 2 Sequences and Series 119 2 6. Find the sum of each of the following series. (a) n ∑ r=1 (3r – 1)2 (b) 2n ∑ r=1 r(r + 2)(r + 5) (c) 2n ∑ r=n r(2r + 3) (d) 3n ∑ r=1 r(r 2 + 1) 7. Using the identity r 4 – (r – 1)4 ≡ 4r 3 – 6r 2 + 4r – 1 and n ∑ r=1 r 2 = 1 6 n(n + 1)(2n + 1), find n ∑ r=1 r 3 . 8. Using the method of differences, find the sum of the first n terms of the series whose rth term, ur , are as follows. (a) ur = 1 (r + 1)(r + 2) (b) ur = 2 (r + 2)(r + 3) (c) ur = 1 r(r + 3) (d) ur = 2r – 1 r(r + 1)(r + 2) 9. Find the sum to n terms of the series. (a) 1 · 52 + 5 · 92 + 9 · 132 + 13 · 172 + … (b) 1 1 · 3 · 4 + 1 2 · 4 · 5 + 1 3 · 5 · 6 + 1 4 · 6 · 7 + … 10. If f(r) = 1 r! , simplify f(r) – f(r + 1). Hence, find n ∑ r=1 r (r + 1)! . 11. If n ∑ r=1 ur = 3n2 + 2n, find ur . Hence, find 2n ∑ r=n+1 ur . 12. If f(r) = 1 r 2 , simplify f(r) – f(r + 1). Hence, find the sum of the first n terms of the series 3 12 · 22 + 5 22 · 32 + 7 32 · 42 + … 13. Find the numbers A, B and C such that 1 + r 2 ≡ A(r + 2)(r + 1) + B(r + 1) + C for all values of r. Hence, prove that n ∑ r=1 (1 + r 2 ) r! = n(n + 1)! 2.3 Binomial Expansions The 1 n r 2 and n! notation Consider the product of the first 50 positive integers as follows: 50 × 49 × 48 × … × 3 × 2 × 1. The product of these integers is a huge number. To simplify and for ease of writing, the product of these numbers may be written as 50! (which is read as “factorial fifty”). Hence, 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3 628 800
120 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 In general, n! is defined as n! = n(n – 1)(n – 2) … 3 · 2 · 1, with n Z+ , and 0! = 1. Notice that (n + 1)! = (n + 1)[n(n – 1)(n – 2) … 3 · 2 · 1] = (n + 1)·n! Example 29 Evaluate 10! 8! . Solution: 10! 8! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 10 × 9 = 90 Alternative method: 10! 8! = 10 × 9! 8! = 10 × 9 × 8! 8! = 90 Example 30 Evaluate 18! 15! 4! . Solution: 18! 15! 4! = 18 × 17 × 16 × 15! 15! × 4! = 18 × 17 × 16 4 × 3 × 2 × 1 = 204 Example 31 Rewrite each of the following products in factorial form. (a) 45 × 44 × 43 × 42 × 41 (b) 10 × 9 × 8 5 × 4 × 3 Solution: (a) 45 × 44 × 43 × 42 × 41 = 45 × 44 × 43 × 42 × 41 × 40 × 39 × … × 3 × 2 × 1 40 × 39 × … × 3 × 2 × 1 = 45! 40! (b) 10 × 9 × 8 5 × 4 × 3 = 10 × 9 × 8 5 × 4 × 3 × 7! 7! × 2! 2! = (10 × 9 × 8 × 7!) × 2! (5 × 4 × 3 × 2!) × 7! = 10! 2! 5! 7! By using (n + 1)! = (n + 1) × n!
Mathematics Term 1 STPM Chapter 2 Sequences and Series 121 2 In problems on permutations and combinations, the number of combinations of choosing r objects from n different objects is given by 1 n r 2. For example, the number of ways of choosing 3 books out of 10 different books is 1 10 3 2, where 1 10 3 2 = 10! 3!7! = 120 ways The notation 1 n r 2 is called the binomial coefficient as it is found in binomial expansions. The binomial coefficient 1 n r 2 is defined as 1 n r 2 = n! (n – r)! r! for n, r Z+ and 0 r n. Notice that 1 n 0 2 = n! (n – 0)! 0! = 1 since 0! = 1 1 n 1 2 = n! (n – 1)! 1! = n(n – 1)! (n – 1)! = n 1 n 2 2 = n! (n – 2)! 2! = n(n – 1)(n – 2)! (n – 2)! 2! = n(n – 1) 2! 1 n 3 2 = n! (n – 3)! 3! = n(n – 1)(n – 2)(n – 3)! (n – 3)! 3! = n(n – 1)(n – 2) 3! So, in general, 1 n r 2 = n! (n – r)! r! = n(n – 1)(n – 2)(n – 3) … (n – r + 1) r! , r n. Example 32 Show that (a) 1 n r 2 = 1 n n – r 2 (b) 1 n r 2 + 1 n r + 1 2 = 1 n + 1 r + 1 2 Solution: (a) From the definition, 1 n r 2 = n! (n – r)! r! and 1 n n – r 2 = n! [n – (n – r)]! (n – r)! = n! r! (n – r)! = n! (n – r)! r! Hence, 1 n r 2 = 1 n n – r 2. (b) From the definition, 1 n r 2 + 1 n r + 1 2 = n! (n – r)! r! + n! [n – (r + 1)]! (r + 1)! = n! (n – r)(n – r – 1)! r! + n! (n – r – 1)! (r + 1) r! = n! (n – r – 1)! r! 3 1 n – r + 1 r + 1 4
122 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 = n! (n – r – 1)! r! 3 r + 1 + n – r (n – r)(r + 1) 4 = (n + 1) n! (n – r)(n – r – 1)! (r + 1)r! = (n + 1)! (n – r)! (r + 1)! 1 n + 1 r + 1 2 = (n + 1)! [(n + 1) – (r + 1)]! (r + 1)! = (n + 1)! (n – r)! (r + 1)! Hence, 1 n r 2 + 1 n r + 1 2 = 1 n + 1 r + 1 2 Both the above results (a) and (b) are very useful in evaluating 1 n r 2, especially when the difference between n and r is small. For any positive integers n and r, n r, (a) 1 n r 2 = 1 n n – r 2 (b) 1 n r 2 + 1 n r + 1 2 = 1 n + 1 r + 1 2 Example 33 Evaluate (a) 1 7 3 2 (b) 1 18 15 2 (c) 1 15 4 2 + 1 15 5 2 (d) 1 12 4 2 + 1 12 7 2 Solution: (a) 1 7 3 2 = 7! (7 – 3)! 3! = 7! 4! 3! = 7 × 6 × 5 × 4! 4! × 3 × 2 × 1 = 35 (b) 1 18 15 2 = 18! (18 – 15)! 15! = 18 × 17 × 16 × 15! 3! × 15! = 18 × 17 × 16 3 × 2 × 1 = 816 (c) 1 15 4 2 + 1 15 5 2 = 1 16 5 2 = 16! 11! 5! = 16 × 15 × 14 × 13 × 12 × 11! 11! × 5 × 4 × 3 × 2 × 1 = 4368 By using the result 1 n r 2 + 1 n r + 1 2 = 1 n + 1 r + 1 2
Mathematics Term 1 STPM Chapter 2 Sequences and Series 123 2 (d) 1 12 4 2 + 1 12 7 2 = 1 12 4 2 + 1 12 12 – 72 = 1 12 4 2 + 1 12 5 2 = 1 13 5 2 = 13! 8! 5! = 13 × 12 × 11 × 10 × 9 × 8! 8! × 5 × 4 × 3 × 2 × 1 = 1287 Exercise 2.6 1. Evaluate (a) 5! (b) 8! (c) 10! (d) 8! 5! (e) 10! 7! (f) 18! 15! (g) 20! 17! 4! (h) 16! 10! 6! (i) 9! 2! 3! 4! (j) 10! (5!) 2. Rewrite in factorial form (a) 10 × 9 × 8 (b) 12 × 11 (c) 22 × 21 × 20 (d) 18 × 17 × 16 5 × 4 × 3 (e) 21 × 20 × 19 8 × 7 × 6 (f) n(n – 1)(n – 2) (g) (n + 2)(n + 1)n(n – 1) (h) (2n + 5)(2n + 4)(2n + 3) (i) n(n – 1)(n – 2) 3 × 2 × 1 (j) (n – 5)(n – 6)(n – 7)(n – 8) 5 × 4 × 3 3. Factorise (a) 10! + 9! (b) 2(8!) – 7! (c) 3(7!) + 4(9!) (d) (n + 1)! + n! (e) (n + 1)! – (n – 1)! (f) n2 (n + 1)! + 2n(n!) (g) (n + 1)! + n! + (n – 1)! (h) 8! 4! 5! + 7! 3! 2! (i) 10! 7! 3! – 8! 5! 4! + 7! 3! 2! (j) n! r! + (n – 1)! (r + 1)! 4. Evaluate (a) 1 8 4 2 (b) 1 10 3 2 (c) 1 20 17 2 (d) 1 15 0 2 (e) 1 6 3 2 1 5 2 2 (f) 1 8 5 2 1 5 2 2 (g) 1 10 4 2 ÷ 1 8 3 2 (h) 1 13 3 2 ÷ 1 12 4 2 (i) 1 16 5 2 + 1 16 12 2 (j) 1 10 4 2 – 1 9 4 2
124 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 The binomial theorem When a binomial expression such as (a + b) is raised to the power of n (where n is a small positive integer), we obtain a binomial expansion such as follows: When n = 0, (a + b) 0 = 1 n = 1, (a + b) 1 = a + b n = 2, (a + b) 2 = a2 + 2ab + b2 n = 3, (a + b) 3 = a3 + 3a2 b + 3ab2 + b3 n = 4, (a + b) 4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 n = 5, (a + b) 5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5 The coefficients of each binomial expansion form an array which is known as a Pascal triangle, as shown below: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Each number in the Pascal triangle is the sum of two numbers adjacent to it in the previous line. For example, 10 = 4 + 6. This means that the triangle can be extended to obtain the coefficients of the binomial expansion of (a + b) n for higher values of n. This way of expansion is rather tedious. An alternative method, known as the binomial theorem, can be used to expand (a + b) n for any n Z+ . Notice that we can arrange the above binomial expansions in the following way: (a + b) 3 = a3 + 3a2 b + 3 · 2 1 · 2 ab2 + b3 (a + b) 4 = a4 + 4a3 b + 4 · 3 1 · 2 a2 b2 + 4 · 3 · 2 1 · 2 · 3 ab3 + b4 (a + b) 5 = a5 + 5a4 b + 5 · 4 1 · 2 a3 b2 + 5 · 4 · 3 1 · 2 · 3 a2 b3 + 5 · 4 · 3 · 2 1 · 2 · 3 · 4 ab4 + b5 In general, if n is a positive integer, we have (a + b) n = an + nan – 1b + n(n – 1) 1 · 2 an – 2b2 + n(n – 1)(n – 2) 1 · 2 · 3 an – 3 b3 + … + bn . However, since n = 1 n 1 2, n(n – 1) 1 · 2 = 1 n 2 2, n(n – 1)(n – 2) 1 · 2 · 3 = 1 n 3 2, … we can expand (a + b) n by using the 1 n r 2 notation, i.e. (a + b) n = an + 1 n 1 2an – 1b + 1 n 2 2an – 2 b2 + 1 n 3 2an – 3 b3 + … + 1 n r 2an – r br + … + 1 n n –1 2abn – 1 + bn , or in short, (a + b) n = n ∑ r=0 1 n r 2an – r br , n Z+ . The above result is called the binomial theorem.
Mathematics Term 1 STPM Chapter 2 Sequences and Series 125 2 The expansion of (1 + x) n , n Z+ From the binomial theorem for (a + b) n , when n is a positive integer with a = 1 and b = x, we have a finite series (1 + x) n = 1 + 1 n 1 2 x + 1 n 2 2 x2 + … + 1 n r 2 xr + … + 1 n n –1 2xn – 1 + xn = 1 + nx + n(n – 1) 2! x2 + … + n(n – 1) … (n – r + 1) r! xr + … + nxn – 1 + xn . In general, if we want to expand (a + b) n , it would be easier if we change it to the form (1 + x) n as shown below: (a + b) n = 3a11 + b a 24n = an 11 + b a 2 n = an (1 + x) n , where x = b a . Example 34 Use the binomial theorem to expand (1 + x) 7 . Solution: (1 + x) 7 = 1 + 1 7 1 2x + 1 7 2 2x2 + 1 7 3 2 x3 + 1 7 4 2x4 + 1 7 5 2x5 + 1 7 6 2x6 + x7 = 1 + 7! 6! 1! x + 7! 5! 2! x2 + 7! 4! 3! x3 + 7! 3! 4! x4 + 7! 2! 5! x5 + 7! 1! 6! x6 + x7 = 1 + 7x + 7 · 6 1 · 2 x2 + 7 · 6 · 5 1 · 2 · 3 x3 + 7 · 6 · 5 1 · 2 · 3 x4 + 7 · 6 1 · 2 x5 + 7x6 + x7 = 1 + 7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7 Example 35 Using the binomial theorem, find the expansion of (3x + 4y) 4 . Solution: (3x + 4y) 4 = (3x) 4 + 1 4 1 2(3x) 3 (4y) + 1 4 2 2(3x) 2 (4y) 2 + 1 4 3 2(3x)(4y) 3 + (4y) 4 = 34 x4 + 4·33 ·4x3 y + 6·32 ·42 x2 y2 + 4·3·43 xy3 + 44 y4 = 81x4 + 432x3 y + 864x2 y2 + 768xy3 + 256y4 Example 36 Find the coefficient of x4 in the expansion of (2x – 1)15. Solution: From the binomial theorem, (a + b) n = n ∑ r=0 1 n r 2an – r br where the (r + 1)th term = 1 n r 2an – rbr Thus (2x – 1)15 = 15 ∑ r=0 1 15 r 2(2x) 15 – r (–1)r The term in x4 is when 15 – r = 4, i.e. r = 11. Coefficient of x4 = 1 15 11 224 (–1)11 = – 15 × 14 × 13 × 12 1 × 2 × 3 × 4 × 24 = –21 840
126 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 Example 37 Expand (1 + x + 2x2 ) 7 up to the term in x3 . Solution: Rewriting (1 + x + 2x2 ) as a binomial, (1 + x + 2x2 ) 7 = [1 + (x + 2x2 )]7 = 1 + 7(x + 2x2 ) + 7 · 6 1 · 2 (x + 2x2 ) 2 + 7 · 6 · 5 1 · 2 · 3 (x + 2x2 ) 3 + … We stop the expansion here because the next term (x + 2x2 ) 4 contains x4 and higher powers of x. = 1 + 7(x + 2x2 ) + 21(x2 + 4x3 ) + 35x3 + … = 1 + 7x + 35x2 + 119x3 + … Ignore x4 and higher powers of x. Exercise 2.7 1. Use the binomial theorem to expand (a) (p + q) 4 (b) (m – n) 7 (c) (2 + k2 ) 6 (d) (2a – b) 5 (e) 1x + 1 x 2 3 (f) 1y – 1 2y 2 8 2. In each of the following expansions, find the term as stated. (a) (1 + x) 10, 5th term (b) (2 – 3x) 8 , term in x2 (c) (2a + b) 12, 10th term (d) (p – 3q2 ) 7 , term in p4 q6 (e) 1x – 1 x 2 6 , constant term (f) 1x2 + 1 x 2 9 , term in 1 x3 . 3. Expand the expression 12x + 1 x2 2 5 + 12x – 1 x2 2 5 , simplifying the terms. 4. The coefficient of x3 is four times the coefficient of x2 in the expansion of (1 + x) n . Find the value of n. 5. In the binomial expansion of 11 + 1 3 x2 n , the coefficients of the fourth and fifth terms are equal. Find the value of n. 6. The coefficient of x5 in the binomial expansion of (1 + 5x) 8 is the same as the coefficient of x4 in the expansion of (a + 5x) 7 . Find the value of a. 7. If the first three terms in the expansion of (1 + ax) n in ascending powers of x are 1 – 4x + 7x2 , find n and a. 8. Find the first four terms of each of the following expansions, in ascending powers of x. (a) (1 + x) 7 (b) (1 + x – x2 ) 7 9. Expand (1 + 2x + 3x2 ) 8 in ascending powers of x up to and including the term in x3 . 10. Find the first three terms, in ascending powers of x, of the expansion (1 – 3x)(1 + 2x) 6 . 11. Find the coefficient of the terms in x as indicated, in the following expansions. (a) (1 + x2 )(2 – 3x) 7 , term in x3 (b) (1 – 3x – 2x2 )(1 + x2 ) 20, term in x20 (c) x1x – 2 x2 2 12 , term in x4 (d) 1x + 1 x 2 2 (1 – x) 5 , term in x2
Mathematics Term 1 STPM Chapter 2 Sequences and Series 127 2 The expansion of (1 + x) n , n Q We have seen that the binomial expansion of (1 + x) n , where n ∈ Z+ , is a finite series with (n + 1) terms, i.e. (1 + x) n = 1 + nx + n(n – 1) 2! x2 + n (n – 1)(n – 2) 3! x3 + … + xn . However, if n is any rational number, i.e. n ∈ Q, then the expansion is an infinite series, i.e. (1 + x) n = 1 + nx + n(n – 1) 2! x2 + n(n – 1)(n – 2) 3! x3 + … + n(n – 1) … (n – r + 1) r! xr + … This series is called the binomial series, and is valid if |x| , 1, i.e. –1 , x , 1. It is used to find approximations, up to a degree of accuracy very close to its actual value. Note: The notation 1 n r 2 = n! (n – r)! r! is not applicable if n is not a positive integer. Example 38 Expand each of the following expressions as an ascending series in x, up to the term in x4 . State the range of x such that the expansion is valid. (a) 11 + 1 2 x2 —1 3 (b) (1 – 2x) –3 Solution: By using the binomial expansion (1 + x) n = 1 + nx + n(n – 1) 2! x2 + n(n – 1)(n – 2) 3! x3 + n(n – 1)\(n – 2)(n – 3) 4! x4 + … (a) Substitute x = 1 2 x and n = 1 3 , we get 1 3 1 1 3 – 12 1 3 1 1 3 – 121 1 3 – 22 11 + 1 2 x2 —1 3 = 1 + 1 3 1 1 2 x2 + 1 1 2 x2 2 + 1 1 2 x2 3 2! 3! 1 3 1 1 3 – 121 1 3 – 221 1 3 – 32 + 1 1 2 x2 4 + … 4! 1 3 1– 2 3 2 1 3 1– 2 3 21– 5 3 2 = 1 + 1 6 x + 1 1 4 x2 2 + 1 1 8 x3 2 2 6 1 3 1– 2 3 21– 5 3 21– 8 3 2 + 1 1 16 x4 2 + … 24 = 1 + 1 6 x – 1 36 x2 + 5 648 x3 – 5 1944 x4 + … The expansion is valid if | 1 2 x| , 1, i.e. |x| , 2 or –2 , x , 2.
128 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 (b) Substitute x = –2x and n = –3, we get (1 – 2x) –3 = 1 + (–3)(–2x) + (–3)(–3 – 1) 2! (–2x) 2 + (–3)(–3 – 1)(–3 – 2) 3! (–2x) 3 + (–3)(–3 – 1)(–3 – 2)(–3 – 3) 4! (–2x) 4 + … = 1 + 6x + (–3)(–4) 2 (4x2 ) + (–3)(–4)(–5) 6 (–8x3 ) + (–3)(–4)(–5)(–6) 24 (16x4 ) + … = 1 + 6x + 24x2 + 80x3 + 240x4 + … The expansion is valid if |–2x| , 1, i.e. 2|x| , 1, or |x| , 1 2 or – 1 2 , x , 1 2 . Example 39 Expand (1 + x) 10 up to and including the term in x3 . Hence, obtain an approximation for (1.01)10 and (0.99)10. Solution: (1 + x) 10 = 1 + 10x + 10 · 9 1 · 2 x2 + 10 · 9 · 8 1 · 2 · 3 x3 + … = 1 + 10x + 45x2 + 120x3 + … Substitute x = 0.01 into the expansion, (1 + 0.01)10 = 1 + 10(0.01) + 45(0.01)2 + 120(0.01)3 + … = 1 + 0.1 + 0.0045 + 0.000120 + … = 1.10462 = 1.1046 (correct to 4 decimal places) Substitute x = –0.01 into the expansion, (1 – 0.01)10 = 1 + 10(–0.01) + 45(–0.01)2 + 120(–0.01)3 + … = 1 – 0.1 + 0.0045 – 0.000120 + … = 0.90438 = 0.9044 (correct to 4 decimal places) Note: By using the calculator, we find that (1.01)10 = 1.1046 and (0.99)10 = 0.9044 (both correct to 4 decimal places). This shows that the accuracy of the approximation does not differ much even if we were to expand (1 + x) 10 up to x5 or more. Hence, usually three or four terms should be sufficient for a good approximation.
Mathematics Term 1 STPM Chapter 2 Sequences and Series 129 2 Example 40 Expand 1 + 2x 1 – 2x as a series in ascending powers of x, up to the term in x2 . By substituting x = 1 100 , find an approximation for 51, stating the number of significant figures your result is accurate to. (1 + 2x) —1 2 Solution: 1 + 2x 1 – 2x = (1 – 2x) —1 2 = (1 + 2x) —1 2 (1 – 2x) – —1 2 1 2 1– 1 2 2 = 31 + 1 2 (2x) + (2x) 2 + …4 1·2 1– 1 2 21– 3 2 2 × 31 + 1– 1 2 2(–2x) + (–2x) 2 + …4 1·2 = 11 + x – 1 2 x2 + …211 + x + 3 2 x2 + …2 = 1 + 2x + 2x2 + … Ignore x3 and higher powers of x. When x = 1 100 , substituting into the expansion 1 + 2 100 1 – 2 100 = 1 + 21 1 1002 + 21 1 1002 2 + … 102 98 = 1 + 1 50 + 1 5000 51 49 = 5101 5000 1 7 51 = 5101 5000 51 = 35 707 5000 = 7.1414 (correct to 5 significant figures) Example 41 Expand (1 + x) – —1 4 in ascending powers of x up to the term in x 2 . Prove that 3 2 11 + 1 802 – —1 4 = 5—1 4 . Using your expansion for (1 + x) – —1 4 and x = 1 80 , find an approximation for 5—1 4 , giving your answer correct to five decimal places.
130 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 1– 1 4 21– 5 4 2 Solution: (1 + x) – —1 4 = 1 + 1– 1 4 2x + x 2 + … 1·2 = 1 – 1 4 x + 5 32 x2 – … 3 2 11 + 1 802 – —1 4 = 3 2 1 81 80 2 – —1 4 = 3 2 1 80 812 —1 4 = 3 2 1 5 × 24 34 2 —1 4 = 3 2 · 2 3 (5)—1 4 = 5 —1 4 By using the expansion of (1 + x) – —1 4 and x = 1 80 , 11 + 1 802 – —1 4 = 1 – 1 4 1 1 80 2 + 5 32 1 1 80 2 2 = 1 – 1 320 + 1 40 960 = 40 833 40 960 Hence, 5 —1 4 = 3 2 × 40 833 40 960 = 1.49535 (correct to 5 decimal places) Exercise 2.8 1. Expand each of the following expressions as a series in ascending powers of x up to the term in x3 . (a) (1 + x) –2 (b) (1 – x) –1 (c) 1 (1 + 2x) 3 (d) (2 + x) –2 (e) (3 + 2x) –1 2. Expand each of the following expressions as a series in ascending powers of x up to the term in x3 . State the range of values of x in each case for which the expansion is valid. (a) 11 + 1 2 x2 – —1 2 (b) (1 – 2x) —1 3 (c) (2 + x) –1 (d) 1 1 + x (e) (1 – x) 1 + x (f) x – 1 x + 1 (g) x + 2 1 – 3x (h) 1 (x – 2)(1 + 2x) (i) (1 + x + x2 ) –1
Mathematics Term 1 STPM Chapter 2 Sequences and Series 131 2 3. Use the expansion of (x + y) 4 to evaluate (1.03)4 , correct to four decimal places. 4. Use the expansion of (2 – x) 5 to evaluate (1.98)5 , correct to five decimal places. 5. Obtain the expansion of (1 + 2x) 15 in ascending powers of x up to the term in x3 . Hence, evaluate (1.002)15, correct to five decimal places. 6. Find the first four non-zero terms of the expansion of (1 + 2x2 ) – —1 2 in ascending powers of x. 7. Obtain the first four terms of the expansion of (1 – x) – —1 2 in ascending powers of x. Deduce the value of 0.9, correct to four decimal places. 8. By substituting x = 0.08 into (1 + x) —1 2 and its expansion, find 3 , correct to four significant figures. 9. By substituting x = 1 10 into (1 – x) – —1 2 and its expansion, find 10 , correct to five significant figures. 10. Expand (2 – x) –2 as a series in ascending powers of x, up to the term in x4 . Deduce the value of 1 (1.8)2 , correct to three significant figures. 11. Expand (1 + 2x) —1 2 in ascending powers of x, up to the term in x3 . By substituting x = 1 8 , find an approximation for 5 , giving your answer correct to three decimal places. 12. Expand 1 – 3x 1 + 4x in ascending powers of x, up to the term in x3 . State the range of values of x for which the expansion is valid. 13. Expand (1 – x – 2x2 ) 5 in ascending powers of x, up to the term in x4 . By substituting x = 0.01, estimate the value of (0.9898)5 , correct to six decimal places. Summary 1. A series Sn = u1 + u2 + u3 + … + un is said to be convergent if there exists a finite number a, such that lim n → ∞ Sn = a. A series is said to be divergent if it is not convergent. 2. Sum of the first n positive integers is n ∑ r=1 r = 1 2 n(n + 1) Sum of the squares of the first n positive integers is n ∑ r=1 r 2 = 1 6 n(n + 1)(2n + 1) Sum of the cubes of the first n positive integers is n ∑ r=1 r 3 = 1 4 n2 (n + 1)2 3. For an arithmetic progression with first term a and common difference d, the nth term is un = a + (n – 1) d. Sum of the first n terms is Sn = n 2 [2a + (n – 1)d]
132 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 4. For a geometric progression with first term a and common ratio r, the nth term is un = arn – 1. Sum of the first n terms is Sn = a(r n – 1) r – 1 , for r > 1 and Sn = a(1 – r n ) 1 – r , for r < 1 For |r| , 1 and large values of n such that lim x → ∞ rn → 0, the sum (to infinity) is S∞ = a 1 – r 5. 1 n r 2 = n! (n – r)! r! , n, r Z+ , 0 r n. = n(n – 1)(n – 2) … (n – r + 1) r! 1 n r 2 = 1 n n – r 2 1 n r 2 + 1 n r + 1 2 = 1 n + 1 r + 1 2 6. The binomial theorem (1 + x) n = 1 + 1 n 1 2x + 1 n 2 2x2 + … + 1 n r 2xr + … + 1 n n – 1 2xn – 1 + xn , for n Z+ . (1 + x) n = 1 + nx + n(n – 1) 2! x2 + … + n(n – 1) … (n – r + 1) r! xr + …, for n Q. STPM PRACTICE 2 1. Write down the nth term of the sequence 0, 2, 6, 12, 20, …. Find n, if the nth term is 210. 2. Find the first three terms and the nth term of the series with the sum to n terms, Sn, given by (a) Sn = 4n – 1 n (b) Sn = (n + 1)! 3. Find the value of (a) 30 ∑ r=1 (r + 2) (b) 8 ∑ r=1 2r (c) n ∑ r=1 2r (d) n ∑ r=1 r(r!) 4. Show that n ∑ r=1 (r + 1)2r – 1 = n2n . 5. (a) Prove that n ∑ r=1 r 2 = 1 6 n(n + 1)(2n + 1). (b) If S = 1(n) + 2(n – 1) + 3(n – 2) + … + r(n + 1 – r) + … + n(1) and T = 1(n – 1) + 2(n – 2) + 3(n – 3) + … + r(n – r) + … + (n – 1)(1), where n is a positive integer, show that S + T = 1 6 n(n + 1)(2n + 1) 6. If un = 1 (2n + 1)(2n + 3) , show that un – 1 – un = 4 (2n – 1)(2n + 1)(2n + 3) . Hence, find the sum to n terms of the series 1 1 · 3 · 5 + 1 3 · 5 · 7 + … + 1 (2r – 1)(2r + 1)(2r + 3) + …
Mathematics Term 1 STPM Chapter 2 Sequences and Series 133 2 7. The arithmetic mean of 1 a + b and 1 b + c is 1 a + c . Find, in terms of b, the arithmetic mean of a2 and c 2 . 8. Find the smallest value of n such that the nth term of the arithmetic series 12.0 + 10.7 + 9.4 + 8.1 + … is negative. Find, also, the smallest value of n such that the sum to n terms is negative. 9. The sum of the first n terms of an arithmetic sequence u1 , u2 , u3 , … is given by Sn = 2n2 + n. Find an explicit formula and recursive formula for un. 10. A sequence is u1 , u2 , u3 , … is defined by un = 4n2 – 1. The difference between successive terms of the sequence forms a new sequence w1 , w2 , w3 , …. (a) Find an explicit formula for wn in terms of n. (b) Show that w1 , w2 , w3 , … forms an arithmetic sequence, and state its first term and common difference. (c) Find the sum of the first n terms of the sequence w1 , w2 , w3 , … in terms of un and wn. 11. If Sn represents the sum of the first n terms of a geometric progression 1 + 1 2 + 1 1 2 2 2 + … + 1 1 2 2 n – 1 + … and S represents the sum to infinity, find the smallest value of n such that S – Sn , 0.001. 12. A geometric sequence is defined by Ur + 1 = 2 + 3 5 Ur , and U1 = 2. (a) Write down each of the terms U2 , U3 and U4 in the form r – 1 ∑ m =0 21 3 5 2 m , and show that explicit formula for Ur is given by Ur = 531 – 1 3 5 2 r 4. (b) Determine the limit of Ur when r tends to infinity. 13. If S is the sum of the series 1 + 3x + 5x2 + … + (2n + 1)xn , by considering (1 – x)S, show that S = 1 + x – (2n + 3)xn + 1 + (2n + 1)xn + 2 (1 – x) 2 , x ≠ 1. 14. By bracketing the terms into pairs, show that 12 – 22 + 32 – 42 + … + (2n – 1)2 – (2n) 2 = –n(2n + 1). Deduce the sum of the series (a) 12 – 22 + 32 – 42 + … + (2n – 1)2 – (2n) 2 + (2n + 1)2 , (b) 252 – 262 + 272 – 282 + … + 492 – 502 . 15. The series of positive integers is grouped into four as follows: (1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), … Show that the sum of the integers in the kth bracket is 2(8k – 3). If the integers are similarly grouped with m integers in each bracket, find (a) the last term in the nth bracket, (b) the first term in the nth bracket. If Sn represents the sum of the integers in the nth bracket, find an expression for Sn in terms of m and n. Hence, show that Sn, S2n, S3n are in arithmetic progression.
134 Mathematics Term 1 STPM Chapter 2 Sequences and Series 2 16. The r th term, ur , of an infinite series is given by: ur = 1 1 2 2 4r – 1 + 1 1 2 2 4r + 1 (a) Express ur in the form k 24r + 1 ,where k is a constant. (b) Find the sum of the first n terms of the series, and deduce the sum of the infinite series. 17. (a) Find the smallest value of n such that the sum of the first n terms of the geometric progression 1 + 0.99 + (0.99)2 + (0.99)3 + … is more than 75% of the sum to infinity. (b) Find the nth term and the sum of the first n terms of the series. 1 + 11 + 111 + 1111 + 11111 + … 18. By expressing the recurring decimals as the sum of a constant and an infinite geometric series, obtain each of the following decimals as a fraction in its lowest terms. (a) 0.273· 1 · (b) 0.597· 5 · (c) 0.7015· 0 · 1 · 19. The sum of the first n terms of a sequence u1 , u2 , u3 , … is given by Sn = 3 2 [1 – 2–n ]. (a) Show that un = 3(2–(n + 1)). (b) Find un + 1 in terms of un, and deduce that the sequence is a geometric sequence. (c) Determine the sum of the series u1 , u2 , u3 , …. 20. Show that 1 1 – 2x – 2 2 + x = 5x (1 – 2x)(2 + x) . Hence, expand 5x (1 – 2x)(2 + x) in ascending powers of x up to the term in x3 . 21. (a) Expand fully the expansions of ( 5 – 3)4 and ( 5 + 3)4 to evaluate ( 5 – 3)4 + ( 5 + 3)4 . (b) Hence, using the inequality of 2 , 5 , 3 and the result from part (a), show that 751 , ( 5 + 3)4 , 752. 22. Write down the first four terms in ascending powers of x for the expansion of (1 + ax) p . Given that the first three terms of this expansion is 1 + 2x + 11 4 x2 , show that 2pa(pa – a) = 11 and find the value of a and p. State the range of values of x for which the expansion is valid. 23. Show that 2511 – 1 252 2 —1 2 can be expressed in the form n 39 , where n is an integer to be found. Expand (1 – x) —1 2 as a series in ascending powers of x up to the term in x2 . By using the first two terms of the expansion of (1 – x) —1 2 with x = 1 252 , obtain an approximate value for 39 in the form p q , where p and q are integers. 24. Two positive integers, p and q, are connected by p = q + 1. Using the binomial expansion, show that the expression p2n – 2nq – 1 can be divided exactly by q2 for all positive integers n. By choosing suitable values for p and n, show that 316 – 33 can be divided exactly by 4, and hence, show that 315 + 5 can be divided exactly by 4. 25. Use the binomial theorem to expand 1 + x 1 – x as a series in ascending powers of x up to the term in x2 , with |x| , 1. By substituting x = 1 10 into your result, show that 11 is approximately 663 200 .
Mathematics Term 1 STPM Chapter 2 Sequences and Series 135 2 26. Show that the first three non-zero terms in the expansion of 11 – 1 n 2 —1 n in ascending powers of 1 1 n 2 is 1 – 1 1 n 2 2 – 1 2 1 1 n 2 3 , and find the term in 1 1 n 2 4 . By giving n a suitable value, use the first four non zero terms of the series to find the value of (0.9)—1 10 , giving your answer correct to five decimal places. 27. (a) Given that n ∑ r=1 r 2 = 1 6 n(n + 1)(2n + 1), obtain an expression, in its simplest terms, for 2n ∑ r=n+1 (2r – 1)2 . (b) Find ∞ ∑ r=1 = 1 103r , expressing your answer as a fraction in its lowest terms. Hence, express the recurring decimal 0.1· 0 · 8 · as a fraction in its simplest form. 28. (a) Find the sum of the arithmetic progression 1, 4, 7, 10, 13, 16, …, 1000. Now, each third term of the progression, i.e. 7, 16, …, is removed. Find the sum of the remaining terms. (b) The r th term, ur , of a series is given by ur = 1 1 3 2 3r – 2 + 1 1 3 2 3r – 1 Express n ∑ r=1 ur in the form A11 – B 27n 2, where A and B are constants. Find A, B and the sum of the series. 29. (a) Find the binomial expansion of (1 + x) —1 4 , for small values of x, up to the term in x2 , with coefficients in their simplest forms. By substituting x = 1 16 in your expression, show that 4 17 ≈ 8317 4096 . (b) Express (x + 2)5 – (x – 2)5 as a polynomial in x, and hence, find the exact value of ( 5 + 2)5 – ( 5 – 2)5 . By assuming that 0 , 5 – 2 , 1 4 , deduce that the difference between ( 5 + 2)5 and an integer is less than 1 1024 . 30. Express f(x) = x2 + 5x (1 + x)(1 – x) 2 in the form A 1 + x + B 1 – x + C (1 – x) 2 , where A, B and C are constants. If the expansion of f(x) in ascending powers of x is c0 + c1 x + c2 x2 + c3 x3 + … + crxr + …, find c0 , c1 , c2 and show that c3 = 11. Express cr in terms of r. 31. (a) Show that for a fixed number x ≠ 1, 2x2 + 2x3 + 2x4 + … + 2xn is a geometric series, and find its sum in terms of x and n. (b) The series Un(x) is given by Un(x) = x + 3x2 + 5x3 + … + (2n – 1)xn , for x ≠ 1. By considering Un(x) – xUn(x) and using the result from (a), show that Un(x) = x + x2 – (2n + 1)xn + 1 + (2n – 1)xn + 2 (1 – x) 2 Hence, determine the value of 15 ∑ r=1 (2r – 1)3r and deduce the value of 14 ∑ r=1 (2r + 1)3r + 2.
3 CHAPTER 3 MATRICES Subtopic Learning Outcome 3.1 Matrices (a) Identify null, identity, diagonal, triangular and symmetric matrices. (b) Use the conditions for the equality of two matrices. (c) Perform scalar multiplication, addition, subtraction and multiplication of matrices with at most three rows and three columns. (d) Use the properties of matrix operations. (e) Find the inverse of a non-singular matrix using elementary row operations. (f) Evaluate the determinant of a matrix. (g) Use the properties of determinants 3.2 Systems of linear equations (a) Reduce an augmented matrix to row-echelon form, and determine whether a system of linear equations has a unique solution, infinitely many solutions or no solution. (b) Apply the Gaussian elimination to solve a system of linear equations. (c) Find the unique solution of a system of linear equations using the inverse of a matrix. augmented matrix – matriks imbuhan column matrix – matriks lajur determinant – penentu diagonal matrix – matriks pepenjuru Gaussian elimination – penghapusan Gauss identity matrix – matriks identiti inverse matrix – matriks songsang null matrix – matriks nol row matrix – matriks baris row-echelon form – bentuk esilon baris singular matrix – matriks singular Bilingual Keywords
137 3 Mathematics Term 1 STPM Chapter 3 Matrices 3.1 Matrices Matrices Consider a set of numbers arranged in a rectangular array of rows and columns, as shown below: 1 2 5 3 1 –2 6 0 4 –1 –3 2 1 2 The numbers that are arranged in such a rectangular array is called a rectangular matrix, or simply, a matrix. The size, or order, of a matrix is determined by the number of rows and columns of the matrix. In the above matrix, there are 3 rows and 4 columns. We say that the matrix is of order 3 × 4. Hence, 1 5 3 4 3 2 2 4 3 6 2 is a matrix of order 3 × 3, 1 1 2 3 4 5 6 2 is a matrix of order 2 × 3, and 1 1 3 5 2 4 6 2 is a matrix of order 3 × 2. A matrix is usually represented by a bold capital letter. For example, A = 1 5 3 4 3 2 2 4 3 6 2 and B = 1 1 3 5 2 4 6 2 . Each number in a matrix is known as an element or entry. In general, a matrix, A, of order m × n, can be represented by a rectangular array of m rows and n columns, as shown below: a11 a21 ai1 am1 a12 a22 ai2 am2 … … … … a1j a2j aij amj … … … … a1n a2n ain amn m rows n columns The entry aij represents an element in the i th row and j th column. For example, in the above matrix A, a12 is an element in the 1st row and 2nd column, i.e. a12 = 3. Similarly, a31 = 4. A matrix is a rectangular array of numbers arranged in rows and columns. A matrix with m rows and n columns is called a matrix of order m × n. An element in the i th row and j th column of a matrix A is represented by aij.
138 3 Mathematics Term 1 STPM Chapter 3 Matrices The square matrix A matrix that has an equal number of rows and columns is called a square matrix. For example, 1 1 3 2 4 2 is a square matrix of order 2 × 2, and 1 4 5 1 1 7 6 3 2 0 2 is a square matrix of order 3 × 3. The column matrix A column matrix is a matrix that has only one column. For example, 1 1 –2 3 2 is a column matrix of order 3 × 1. The row matrix A row matrix is a matrix that has only one row. For example, (–1 0 2) is a row matrix of order 1 × 3. The null matrix A matrix, of any size, where each element of the matrix is 0 is called a null matrix, and is denoted by 0. The null matrix can be of any order. For example, 1 0 0 0 0 2 is a null matrix of order 2 × 2, whereas 1 0 0 0 0 0 0 2 is a null matrix of order 3 × 2. The identity matrix Suppose that each element on the major diagonal of a diagonal matrix is 1. Then this matrix is known as an identity matrix, represented by I. Hence, I = 1 1 0 0 0 1 0 0 0 1 2 is an identity matrix of order 3 × 3. The diagonal matrix Consider the square matrix M of order 3 × 3, where M = major diagonal 1 2 0 0 0 –1 0 0 0 3 2 Each element in M is 0, except for the elements on the major diagonal. A matrix of this form is known as a diagonal matrix of order 3 × 3. The triangular matrix If the elements below the major diagonal of a square matrix are all zeros, the matrix is called an upper-triangular matrix.
139 3 Mathematics Term 1 STPM Chapter 3 Matrices Hence, P = 1 1 0 4 3 2 and Q = 1 3 0 0 2 5 0 4 –1 6 2 are upper-triangular matrices of order 2 × 2 and 3 × 3 respectively. If the elements above the major diagonal are all zeros, then the matrix is called a lower-triangular matrix. Hence, X = 1 2 5 0 –32 and Y = 1 4 6 –2 0 4 2 0 0 5 2 are lower-triangular matrices of order 2 × 2 and 3 × 3 respectively. The symmetric matrix Consider the elements in the following square matrix of order 3 × 3, i.e. S = 1 1 3 4 3 2 1 4 1 –1 2 major diagonal (axis of symmetry) When the matrix is reflected about the major diagonal, we notice that the matrix remains unchanged. We say that this 3 × 3 matrix is a symmetric matrix. Note: A diagonal matrix, an identity matrix, and a symmetric matrix are all square matrices. Equal matrices Two matrices are equal if the order of the matrices are the same and each corresponding element in the two matrices are equal. For example, if A = 1 a 2 5 0 1 6 2 and B = 1 4 b 5 0 1 c 2 and it is given that A = B, then 1 a 2 5 0 1 6 2 = 1 4 b 5 0 1 c 2 Both matrices are of order 3 × 2, and corresponding elements are equal and a = 4, b = 2 and c = 6. But 1 5 0 6 7 4 3 2 ≠ 1 5 7 0 4 6 3 2 The orders are not the same Order 3 × 2 Order 2 × 3 Addition and subtraction of matrices If A and B are two matrices of the same order, we can perform simple operations such as the addition or subtraction of matrices A and B, written as A + B and A – B. This addition or subtraction of matrix A and matrix B can be performed by adding or subtracting the corresponding elements in matrix A and matrix B. For example, if A = 1 5 3 4 3 2 2 4 3 6 2 and B = 1 6 4 5 5 3 4 5 7 8 2
140 3 Mathematics Term 1 STPM Chapter 3 Matrices then A + B = 1 5 3 4 3 2 2 4 3 6 2 + 1 6 4 5 5 3 4 5 7 8 2 = 1 5 + 6 3 + 4 4 + 5 3 + 5 2 + 3 2 + 4 4 + 5 3 + 7 6 + 82 = 1 11 7 9 8 5 6 9 10 14 2 and B – A = 1 6 4 5 5 3 4 5 7 8 2 – 1 5 3 4 3 2 2 4 3 6 2 = 1 6 – 5 4 – 3 5 – 4 5 – 3 3 – 2 4 – 2 5 – 4 7 – 3 8 – 6 2 = 1 1 1 1 2 1 2 1 4 2 2 This addition or subtraction can be extended to three or more matrices. We know that the addition operation on real numbers obey the commutative and associative laws, i.e. for any a, b, c R, a + b = b + a and (a + b) + c = a + (b + c). Similarly, the addition operation on two or more matrices of the same order also obey the commutative and associative laws. For any matrices A, B and C of the same order, A + B = B + A – Commutative rule of addition (A + B) + C = A + (B + C) – Associative rule of addition Multiplication of a matrix by a scalar When a matrix A, is multiplied by a scalar, k, it means that each element in matrix A is multiplied by k. For example, if A = 1 5 3 4 3 2 2 4 3 6 2 , then kA = k 1 5 3 4 3 2 2 4 3 6 2 = 1 5k 3k 4k 3k 2k 2k 4k 3k 6k 2 and 2A = 2 1 5 3 4 3 2 2 4 3 6 2 = 1 10 6 8 6 4 4 8 6 12 2 If k = 0, then kA = O, i.e. the result is a null matrix.
141 3 Mathematics Term 1 STPM Chapter 3 Matrices Multiplication of two matrices If A is a matrix of order m × p and B is a matrix of order p × n, it is possible to perform the operation of multiplication of matrix A and matrix B written as AB. The product AB is a new matrix of order m × n. Each element of the i th row and j th column in the matrix AB, cij, is obtained by multiplying the elements of the i th row of matrix A with the corresponding elements of the jth column of matrix B and the results added up, i.e. jth column jth column cij A B AB = i th row i th row The multiplication of two matrices, A and B, is possible if and only if the number of columns of matrix A is equal to the number of rows of matrix B. If A is a matrix of order m × p and B is a matrix of order p × n, the product of the matrices, AB, exists and is a matrix of order m × n. Example 1 If A = 1 1 –2 3 2 2 and B = 1 –2 4 3 0 1 –3 2, determine if each of the products AB and BA exists. If the product exists, find the resulting matrix. Solution: (a) The order of matrix A is 2 × 2, and the order of matrix B is 2 × 3. The number of columns of matrix A is equal to the number of rows of matrix B, i.e. 2. Hence, the product AB exists and is of order 2 × 3. AB = 1 1 –2 3 2 21 –2 4 3 0 1 –3 2 = 1 –2 + 12 4 + 8 3 + 0 –6 + 0 1 – 9 –2 – 6 2 = 1 10 12 3 –6 –8 –8 2 (b) For the product BA, since B is of order 2 × 3 and A is of order 2 × 2, the number of columns of B ≠ the number of rows of A. Hence, the product BA does not exist. The matrix An , where n Z+ For any matrix A, the product AA is only possible if the number of rows and number of columns of A are equal, i.e. AA exists only if A is a square matrix. We can write A2 = AA. For example, if A = 1 1 2 3 4 2, A2 = AA = 1 1 2 3 4 21 1 2 3 4 2 = 1 7 10 15 22 2
142 3 Mathematics Term 1 STPM Chapter 3 Matrices Now, AA2 = 1 1 2 3 4 21 7 10 15 22 2 = 1 37 54 81 1182 Similarly, A2 A = 1 7 10 15 22 21 1 2 3 4 2 = 1 37 54 81 1182 Hence, we see that AAA = AA2 = A2 A or A3 = AAA This rule can be extended for any square matrix A and n a positive integer, i.e. An = 14243 AAA … A n times If A is a square matrix and n a positive integer, then An = AAA … A (n times). Example 2 Find (a) 1 2 –1 3 2(6 2 5) (b) 1 1 0 3 2 –1 5 –3 4 2 21 2 4 1 –3 5 2 1 –2 3 2 Solution: (a) 1 2 –1 3 2(6 2 5) = 1 2 × 6 –1 × 6 3 × 6 2 × 2 –1 × 2 3 × 2 2 × 5 –1 × 5 3 × 52 = 1 12 –6 18 4 –2 6 10 –5 15 2 (b) 1 1 0 3 2 –1 5 –3 4 2 21 2 4 1 –3 5 2 1 –2 3 2 = 1 1 × 2 + 2 × 4 + (–3) × 1 0 × 2 + (–1) × 4 + 4 × 1 3 × 2 + 5 × 4 + 2 × 1 1 × (–3) + 2 × 5 + (–3) × 2 0 × (–3) + (–1) × 5 + 4 × 2 3 × (–3) + 5 × 5 + 2 × 2 1 × 1 + 2 × (–2) + (–3) × 3 0 × 1 + (–1) × (–2) + 4 × 3 3 × 1 + 5 × (–2) + 2 × 3 2 = 1 7 0 28 1 3 20 –12 14 –1 2 Note: If A is a matrix of order m × n and B is a matrix of order n × m (m ≠ n), then AB is a square matrix of order m × m, whereas BA is a square matrix of order n × n. Hence, AB ≠ BA.
143 3 Mathematics Term 1 STPM Chapter 3 Matrices Example 3 If A = 1 2 4 3 –2 –1 5 2 and B = 1 4 –3 1 0 2 –2 2, show that AB ≠ BA. Solution: AB = 1 2 4 3 –2 –1 5 21 4 –3 1 0 2 –2 2 = 1 2 × 4 + 3 × (–3) + (–1) × 1 4 × 4 + (–2) × (–3) + 5 × 1 2 × 0 + 3 × 2 + (–1) × (–2) 4 × 0 + (–2) × 2 + 5 × (–2)2 = 1 –2 27 8 –142 BA = 1 4 –3 1 0 2 –2 21 2 4 3 –2 –1 5 2 = 1 4 × 2 + 0 × 4 –3 × 2 + 2 × 4 1 × 2 + (–2) × 4 4 × 3 + 0 × (–2) –3 × 3 + 2 × (–2) 1 × 3 + (–2) × (–2) 4 × (–1) + 0 × 5 –3 × (–1) + 2 × 5 1 × (–1) + (–2) × 52 = 1 8 2 –6 12 –13 7 –4 13 –11 2 From the above results, we see that AB is of order 2 × 2, whereas BA is of order 3 × 3. Hence, AB ≠ BA. Example 4 Express as a single matrix. (a) 1 3 –2 1 7 4 –3 2 + 41 –1 2 5 –2 6 0 2 (b) 1 1 0 0 1 –1 1 21 3 2 –1 1 –2 2 21 –1 1 2 Solution: (a) 1 3 –2 1 7 4 –3 2 + 41 –1 2 5 –2 6 0 2 = 1 3 –2 1 7 4 –3 2 + 1 –4 8 20 –8 24 0 2 = 1 3 – 4 –2 + 8 1 + 20 7 – 8 4 + 24 –3 + 0 2 = 1 –1 6 21 –1 28 –3 2 (b) 1 1 0 0 1 –1 1 21 3 2 –1 1 –2 2 21 –1 1 2 = 1 3 + 0 + 1 0 + 2 – 1 1 + 0 – 2 0 – 2 + 2 21 –1 1 2 = 1 4 1 –1 0 21 –1 1 2 = 1 –4 – 1 –1 + 0 2 = 1 –5 –1 2
144 3 Mathematics Term 1 STPM Chapter 3 Matrices Alternatively, we have 1 1 0 0 1 –1 1 21 3 2 –1 1 –2 2 21 –1 1 2 = 1 1 0 0 1 –1 1 21 –3 + 1 –2 – 2 1 + 2 2 = 1 1 0 0 1 –1 1 21 –2 –4 3 2 = 1 –2 + 0 – 3 0 – 4 + 3 2 = 1 –5 –1 2 Hence, ABC = (AB)C = A(BC) The product of three matrices A, B and C obeys the associative law, i.e. ABC = (AB)C = A(BC) Example 5 If A = 1 2 –4 –1 3 1 0 1 2 5 2 and I = 1 1 0 0 0 1 0 0 0 1 2, find (a) AI, (b) IA. Hence, state a characteristic for the matrix I. Solution: (a) AI = 1 2 –4 –1 3 1 0 1 2 5 21 1 0 0 0 1 0 0 0 1 2 = 1 2 + 0 + 0 –4 + 0 + 0 –1 + 0 + 0 0 + 3 + 0 0 + 1 + 0 0 + 0 + 0 0 + 0 + 1 0 + 0 + 2 0 + 0 + 5 2 = 1 2 –4 –1 3 1 0 1 2 5 2 = A (b) IA = 1 1 0 0 0 1 0 0 0 1 21 2 –4 –1 3 1 0 1 2 5 2 = 1 2 + 0 + 0 0 – 4 + 0 0 + 0 – 1 3 + 0 + 0 0 + 1 + 0 0 + 0 + 0 1 + 0 + 0 0 + 2 + 0 0 + 0 + 5 2 = 1 2 –4 –1 3 1 0 1 2 5 2 = A From the above results, we see that for any square matrix A, of order m × m, the product of A with an identity matrix I, of the same order m × m, results in the matrix A, i.e. AI = IA = A