1 CHAPTER 3 SEQUENCES AND SERIES 3.1 SEQUENCES AND SERIES 3.2 BINOMIAL EXPANSION
2 CHAPTER 3 SEQUENCES AND SERIES 3.1 SEQUENCES AND SERIES LECTURE 1 OF 2
series OBJECTIVES At the end of this topic, students should be able to. Write n th term of simple sequences and series Find the n th term of arithmetic sequence and series, Tn a (n1)d and use the sum formula, or 2 1 2 ( ) n n S a n d 2 n n S a l (a) (b)
series OBJECTIVES At the end of this topic, students should be able to. (c) (d) Find the n th term of Geometric Sequences and Series, and use the sum formula, 1 Tn n ar1 1 1 Sn for n a r r r Find the Sum To Infinity 1 1 , where a S r r
2, 4, 6, 8, 10 2 + 4 + 6 +8 + 10 (has the last term ) 1, 3, 5, 7, ... 1 + 3 + 5 + 7 + ... (no last term) Finite sequence Finite series Infinite sequence Infinite series 4 10 6 2 8 SEQUENCES AND SERIES
Find the n th term of the sequence 4, 7, 10, 13, 16, … Tn = 3n + 1 T1 = 4 = 3(1) + 1 T2 = 7 = 3(2) + 1 T3 = 10 = 3(3) + 1 . . T5 = 16 = 3(5) + 1 Example 1
ARITHMETIC SEQUENCE An arithmetic sequence with the first term, a and the common difference, d is of the form a, a + d, a + 2d, a + 3d, … The n th term of an arithmetic sequence is given by Tn = a + ( n – 1 )d T1 = a + 0d = a + (1 – 1)d T2 = a + 1d = a + (2 – 1)d T3 = a + 2d = a + (3 – 1)d T4 = a + 3d = a + (4 – 1)d . . . Tn = a + (n – 1)d
The 11th term of an arithmetic sequence is 52 and the 19th term is 92. Find the 1000th term. Tn a (n1)d T11 a 10d 52 T19 a 18d 92 (i) (ii) Example 2
1000 T 2 (1000 1)(5) 4997 (ii) – (i), 8d = 40 d = 5 a = 2
SUM OF A FINITE ARITHMETIC SERIES M S U ( 1) 2 n n S a a n d a l n Sn 2 a n d n Sn 2 ( 1) 2 Let the last term, l = a +(n – 1)d
In an arithmetic sequence, the 10th term is 3 and the sum for the first six terms is 76.5. Find (a) the first term and the common difference, (b) the number of terms to be taken so that its sum is zero. Example 3
.…….. (1) 6 (2 5 ) 76.5 2 S6 76.5: a d T10 3: a d 9 3 6 15 76.5 a d ... (2) 6(3 9 ) 15 76.5 d d 18 54 15 76.5 d d 39 58.5 d d 1.5 a 16.5
13 2 16 5 1 1 5 0 2 ( . ) ( )( . ) n n S n 3 33 1 0 2 n n( ) n n ( ) 66 3 3 0 n n ( ) 69 3 0 n n 0 23 ,
A Geometric Sequence is a sequence of numbers in which each term can be obtained from the previous term by multiplying by a constant number called the common ratio. a , a r , a r2 , a r3 , . . . Let a be the first term and r be the common ratio. Geometric Sequence 1 1 1 T a ar 2 1 2 T ar ar 2 3 1 3 T ar ar 3 4 1 4 T ar ar . . . 1 n T ar n
1 n Tn ar r The nth term is The common ratio, 2 1 T T 3 2 T T ... 1 n n T T M S U
1 n T ar n a = 3 , Given sequence 3, -6, 12, -24, … . Find (a)the nth term (b) the 9th term If the kth term is 192, find the value of k. (a) 6 2 3 r Example 4
T9 = 3(-2)8 = 768 1 3( 2) 192 k Tk 1 ( 2) 64 k 1 6 ( 2) ( 2) k k 1 6 k 7
The Sum of Geometric Series (1 ) 1 1 n n a r S for r r
Example 5 (a) the sum of the first 5 terms, (b) the least number of terms the sequence can have if the sum exceeds 550. The first term of a geometric sequence is 27 4 and the common ratio is . Find 3
(a) a = 27, r = 5 5 4 27 1 3 781 4 3 1 3 S 4 3
4 550 1 3 81 n 4 631 3 81 n Sn > 550 4 27 1 3 550 4 1 3 n 4 631 log log 3 81 n 631 log 81 4 log 3 n n 7.136 Therefore, the least number of terms is 8. 4 log 0 3
Sum To Infinity of a Geometric Series 1 r 1 r 1 1 1 n n a r S r 1 1 , where a S r r , n As n r 0 1 1 n a ar r r
Example 6 Find the sum to infinity for the series 1 3+1+ +... 3 a S 1 r 3 1 1 3 2 9 1 a 3 r 3
Stating a Repeated Decimal Number in the Simplest Fractional form Example 7 Express 0.48 as an infinite geometry series and state as a fraction in the simplest form. 0.48 0.484848484848... = 0.48 + 0.0048 + 0.000048 + …
This is an infinite geometric series with 0.0048 1 a 0.48, r 0.48 100 0.48 1 1 100 100(0.48) 99 0.48 Since -1 < r < 1, the sum to infinite exists. Hence, a S 1 r = 0.48 + 0.0048 + 0.000048 + … 16 33
A lecture hall has 50 rows of seats with 30 seats in the first row, 32 in the second, and 34 in the third and so on. Find the total number of seats. The number of seats in the rows forms an AP, that is: 30, 32, 34, … a = 30, d = 2, S50 = ? 50 50 2 30 49 2 3950 2 S ( ) ( ) Example 8
JOB VACANCY FOR ONE MONTH Job 1 One million dollars at the end of the month. Job 2 Two cents on the first day, 4 cents on the second day, 8 cents on the third day, … Which job will you choose?
Job 1: RM1,000,000 2 3 Job 2: . 2 2 2 2 .. n Sn 30 30 S 2 2 1 ( ) cents How long you need to work for job 2 to earn one million dollar? 2 2 1 2 1 ( ) n 2 2 1 ( ) cents n = RM21,474,836 DAY DAILY PAY 1 0.02 2 0.04 3 0.08 4 0.16 5 0.32 6 0.64 7 1.28 8 2.56 9 5.12 10 10.24 11 20.48 12 40.96 13 81.92 14 163.84 15 327.68 16 655.36 17 1310.72 18 2621.44 19 5242.88 20 10485.76 21 20971.52 22 41943.04 23 83886.08 24 167772.16 25 335544.32 26 671088.64 27 1342177.28 28 2684354.56 29 5368709.12 30 10737418.24 JOB 2 21474836.46
1. Find the n th term of the series 5 + 11 + 21 + 35 + … Exercise Solution Tn = 2n 2 + 3 T1 = 5 = 2(1) 2 + 3 T2 = 11 = 2(2) 2 + 3 T3 = 21 = 2(3) 2 + 3 . . T5 = 53 = 2(5) 2 + 3
The n th term of a sequence is . Identify that the sequence is an arithmetic sequence . 12 4n5 2. 4 5 12 n n T Tn1 12 4 1 4( 1) 5 n 12 n To prove whether a given sequence is an arithmetic sequence: Do not show by calculating and T T 2 1 T T 3 2 T T n n 1 12 4 1 12 4 5 n n 3 1 3 1 Since is a constant, therefore the sequence is an arithmetic sequence. T T n n 1 Solution
Find the number of terms between 100 and 500 that are divisible by 7. Consider : 101, ….., 499 497 105 ( 1)(7) = + - n n 57 105, 112, ….,497 a = 105, d = 7, Tn = 497 ( 1) T a n d n The numbers that are divisible by 7 are Solution 3.
The sum of the first n terms of an arithmetic series is given by Sn = 2n + n 2 . Find (a) an expression for the n th term and the common difference. (b) the sum of the 11th term to the 20th term. 4. 2 2 2 2 1 1 n n n n 2 2 2 2 2 2 1 n n n n n (a) Tn S S n n 1 2 S n n n 2 2n1 2 1 2 1 1 n n ( ) 2 1 2 1 2 n n d T T n n 1 320 2 2 20 20 ( ) ( ) 440 120 S S 20 10 ( ) ... b T T T 11 12 20 2 2 10 10 Solution
5. The sum to infinity of a geometric series is 7 and the sum of the first two terms is Find the positive common ratio and the corresponding first term. 48 . 7 a S 7 1 r a 7 1 r …(1) 2 48 S a ar 7 a 48 1 r 7 …(2) 2 a ar ar ... 2 2 (1 ) 48 1 7 a r S r 48 7(1 r)(1 r)= 7 1 r 7 2 48 1 r = 49 2 1 r = 49 Since r 0 1 r 7 1 a 7 1 6 7 Solution
0.235 = 0.2353535… = 0.2 + 0.035 + 0.00035 + 0.0000035+… infinite geometric series 0.035 1 1 100 S 100 0.035 99 7 198 6. Express as an infinite geometry seri 0.235 es and state as a fraction in the simplest form. = 0.2 + S 0.235 ∞ 7 0.2 198 1 7 5 198 233 990 Solution
In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year. a) Write a formula for the student population in the year. b) Estimate the student population in 2020. th n a) 7 T 284(1.04) 8 374 n 1 n a 284, r 1.04 T 284(1.04) Solution 7. b)
SUM OF A FINITE ARITHMETIC SERIES Sn = a + [a+d] + [a+2d] + … +[a+(n-2)d] +[a+(n-1)d] -----(1) (1) + (2): 2Sn = n terms Reverse the order of addition: Sn = [a+(n-1)d] + [a+(n-2)d] + … + [a+2d] + [a+d] + a -----(2) [2a+(n-1)d] + [2a+(n-1)d] + … + [2a+(n-1)d] + [2a+(n-1)d] EXTRA INPUT FOR YOU…
( 1) 2 n n S a a n d a l n Sn 2 a n d n Sn 2 ( 1) 2 2 Sn= Let the last term, l = a +(n – 1)d n [ 2a +(n-1)d ]
The Sum of Geometric Series n n rSn ar ar ar ar ar 2 3 1 ... 2 2 1 ... n n n S a ar ar ar ar Multiply (1) by r (1) (2)
(1) - (2) S rS n n (1 ) (1 )n S r a r n 2 3 2 1 ... n n Sn a ar ar ar ar ar n n n rSn ar ar ar ar ar ar 2 3 2 1 ... (1) (2) (1 ) 1 1 n n a r S for r r n a ar
Sn = T1 + T2 + T3 +………….. + Tn-1 + Tn ------(1) Tn = Sn - Sn - 1 Relationship Between Tn and Sn (1) – (2) : Sn – Sn-1 = Tn Sn-1 = T1 + T2 + T3 +………….. + Tn-1 ------(2)