Chapter 3 Sequences and Series

Chapter 3 : Sequences And Series Mathematics SM015

Q1 : 2003/2004
Find the sum of even numbers between 199 and 1999.

200, 202, 204, …, 1998
= 200 , = 2
= + ( − 1) = 1998

200 + ( − 1)(2) = 1998
= 900

Therefore, = 900 (2(200) + 899(2)) = 989100

2

Q2 : 2003/2004
The sum of the first four terms of a geometric series with common ratio − 1 is 30.

2
Determine the tenth term and the infinite sum, S .

4 = 30 10 = 9 = 48 (−1)9 = − 3
32
2

(1−(−21)4 = 30
1−(−21)

(1 − 1 ) = 30 × 3 ∞ = = 48 = 32
1− 1−(−21)
16 2

15 = 45

16

= 48

Q3: 2003/2004

The sum of the first n terms of an arithmetic sequence is n (4n + 20)

2

(a) Write down the expression for the sum of the first (n −1)terms.

(b) Find the first term and the common difference of the above sequence.

Chapter 3 : Sequences And Series Mathematics SM015

(a)

Given = ( + ),


Thus −1 = −1 ( ( − 1) + ) = ( −1)(4 +16) = 2( − 1)( + 4)
2
2

(b)
1

= 1 = 1 = 2 (4(1) + 20) = 12
2

2 = 2 − 1 = 2 [4(2) + 20] − 12 = 16
= 2 − 1 = 16 − 12 = 4

Q4: 2004/2005
Express 5.555… in the form of geometric series. Hence, find the

(a) sum of the first n terms.
(b) infinite sum of the series.

5.555… = 5 + 0.5 + 0.05 + 0.005 + …

0.5
= 5, = 5 = 0.1

(a) = (1− ) = 5(1−0.1 ) = 50(1−0.1 )
1− 1−0.1 9

(b) ∞ = 50(1−0.1∞) = 50(1−0) = 50
9 9 9

Q5: 2005/2006
11

The third and the sixth terms of a geometric series are and . Determine the values of
2 16

the first term and the common ratio. Hence, find the sum of the first nine terms of the series.

3 = 2 = 1 ------- (1)
2

6 = 5 = 1 --------(2)

16

Chapter 3 : Sequences And Series Mathematics SM015

(2) : 5 = 1 Thus, (1)2 =1
(1) 2 16
1 2 2

2

3 = 1 = 1

8 42

= 1 = 2

2

9 = (1− 9) = 2(1−(21)9) = 511
1− 1−12 128

Q6: 2005/2006

Expand 1 up to the term x3 and determine the interval of x for which the expansion
(3 − x)3

1
is valid. Hence, approximate (2.9)3 correct to four decimal places.

1 = (3 − )−3
(3 − x)3

= (3 (1 − ))−3

3

= 3−3 (1 − )−3

3

= 1 [1 + (−3) (− ) + (−3)(−3−1) (− )2 + (−3)(−3−1)(−3−2) (− )3 + ⋯ ]

27 3 2! 3 3! 3

= 1 [1 + + 2 2 + 10 3 + ⋯ ]

27 3 27

= 1 + 1 + 2 2 + 10 3 + ⋯
27 27 81 729

The expansion is valid when ⌈− ⌉ < 1

3

⌈ ⌉ < 1

3

−1 < < 1

3

−3 < < 3

Chapter 3 : Sequences And Series Mathematics SM015

1 = (2.9)−3
(2.9)3

Compare with (3 − )−3 = (2.9)−3

3 − = 2.9

= 0.1

Therefore, (2.9)−3 = = 1 + 1 (0.1) + 2 (0.1)2 + 10 (0.1)3 + ⋯ = 0.0410 ( 4 d.p)
27 27 81 729

Q7:2006/2007
The sum of the first k terms of an arithmetic series is 777. The first term is − 3 and the
k-th term is 77. Obtain the value of k and the eleventh term of the series.


= 2 ( + ) = 777

(−3 + 77) = 777

2

= 21

∴ 21 = 77 ∴ 11 = + 10
+ 20 = 77 = −3 + 10(4)
−3 + 20 = 77 = 37
20 = 80
= 4

Q8: 2006/2007

(a) Find the first four terms in the binomial expansion of the following functions:

(i) 1+ 2x 1
(ii) (1 − x)2

(b) Hence, expand 1+ 2x in ascending power of x up to the term containing x3 .
(1 − x)4

By putting x = 1 , show that 12000 is approximately 10947 .
10 100

Chapter 3 : Sequences And Series Mathematics SM015

(a) (i) 1

1 + 2x = (1 + 2 )2

= 1 + 1 (2 ) + ( )( − ) (2 )2 + (21)(12−1)(21−2) (2 )3 + ⋯
2 ! 3!

= 1 + − 1 2 + 1 3 + ⋯
2 2

(ii) 1 = (1 − )−2
(1 − x)2

= 1 + (−2)(− ) + ( )(− − ) (− )2 + (−2)(−2−1)(−2−2) (− )3 + ⋯

! 3!

= 1 + 2 + 3 2 + 4 3+…

(b) 1+ 2x = (1 + 1 − )−2

(1 − x)4 2 )2(1

= (1 + − 1 2 + 1 3 + ⋯ ) (1 + 2 + 3 2 + 4 3 + ⋯ )
2 2

= 1 + 2 + 3 2 + 4 3 + + 2 2 + 3 3 − 1 2 − 3 + 1 3 + ⋯

22

= 1 + 3 + 9 2 + 13 3 + ⋯

22

When = 1
10

√ ( +− 1(11010) ) = 1 + 3 1 + 9 12 + 13 13 + ⋯
(10) 2 (10) 2 (10)

12 2703
= 2000
√ 10
6561

10000

√12 × 10000 = 2703
10 6561 2000

Chapter 3 : Sequences And Series Mathematics SM015

√12000 2703
81 = 2000

10947
√12000 ≈ 100

Q9: 2007/2008
The sum of the first n terms of an arithmetic series is n (3n − 5) . If the second and fourth

2
terms of the arithmetic series are the second and the third terms of a geometric series
respectively, find the sum of the first eleven terms of this geometric series.

Sn = n (3n − 5)
2

T2 = S2 − S1 T4 = S4 − S3

= 2 3(2)− 5− 1 3(1)− 5 = 4 3(4)− 5− 3 3(3)− 5

22 22
= 14 − 6
= (1)− (−1) =8

=2

Tn = arn−1

2 = ar → (1) 8 = ar2 → (2)

(2) (1),  ar2 = 8  r = 4, a = 1
ar1 2 2

( ) 1  411 −1
 2 
 S11 = 4 = 699050.5
−1

Chapter 3 : Sequences And Series Mathematics SM015

Q10: 2008/2009

The fifth term and the tenth term of a geometric series are 3125 and 243 respectively.

(a) Find the value of common ratio, r of the series.

(b) Determine the smallest value of n such that S − Sn  0.02 , where Sn is the
S

sum of the first n term and S is the sum to infinity of the geometric series.

(a) ar9 =243
ar4 = 3125

ar9 = 243
ar 4 3125

r 5 =  3 5
5

r =3
5

(b) S − Sn  0.02
S

∞ − < 0.02

∞ ∞

1− < 0.02

∞

> 0.98

∞

(1− )

1− > 0.98


1−

(1− ) × 1− > 0.98

1−

1− > 0.98
< 0.02
3
(5) < 0.02
log 3 < log 0.02

5

Chapter 3 : Sequences And Series Mathematics SM015

> log 0.02 3
( ℎ ℎ because log 5 = −0.2218)
log 3
5

> 7.66
= 8

Q11: 2008/2009

(a) The rth term of an arithmetic progression is (1+ 6r). Find in terms of n, the sum of the

first n terms of the progression.

(b) (i) Show that 1 = 1 1 − x  − 1
2

9− x 3 9

1 − x  − 1
2
(ii) Find the first three terms in the binomial expansion in ascending power of
 9

x and state the range of the values of x for which this expansion is valid.
(iii) Find the first three terms in the expansion of 3(1+ x) in ascending powers of x

9−x

(a)

1 = (9 − )−21 = [9 (1 − )]−21
√9− 9

= 9−21 (1 − )−12

9

= 1 (1 − 9 )−21 ( ℎ )
3

(b) (1 − )−21 =  1 +   (− 1) (− )   +   (−21)(−12−1) (− )2 +. ..
29 29
9 9 )−12

(1 − = 1  +     + 2 +. ..
18   216

Chapter 3 : Sequences And Series Mathematics SM015

The Validity of the expansion: | | < 1

9

−1 < < 1

9

  −9 < < 9

3(1 + ) 3(1 + )
√9 + = 3(1 + 9 )21

= (1 + ) (1 + 9 )−21

= (1 + ) (1  +     +   2 +. . . )

18 216

= 1  +     +   2 + + 2 +. ..
18 216 18

= 1  +   19   +   213 2+. ..
18 216

Q12: 2009/2010

The first term of common difference of an arithmetic progression are a and − 2 , respectively.
The sum of first n terms is equal to the sum of first 3n terms. Express a in terms of n. Hence,
show that n = 7 if a = 27 .

Sn = n 2a + (n −1)(− 2) = an − n2 + n

2

S3n = n 2a + (3n −1)(− 2) = 3an − 9n2 + 3n

2

Sn = S3n ,

an − n2 + n = 3an − 9n2 + 3n

2an = 8n2 − 2n

 a = 4n −1

i.e 4n −1 = 27 → n=7

Chapter 3 : Sequences And Series Mathematics SM015

Q13: 2009/2010

(a) Expand 1 and −1 x2 .

(4 − x) 2 (1 + 3x) 2 in ascending powers of x up to the terms

1 −1 x2 and determine the range of x

(b) Find the expansion of (4 − x) 2 (1 + 3x) 2 up to the terms

such that this expansion is valid. Hence, by substituting x = 1 , approximate the value of
13

51 correct to four significant figures.

11

(a) (4 − x) 2 = [4 (1 − )]2

4

1

= 2 (1 − )2

4

= 2 [1 + 1 (− ) + (21)(21−1) (− )2 + ⋯ ]
24 2! 4

= 2 [1 − − 2 + ⋯ ]

8 128

= 2 − − 2 + ⋯

4 64

−1 = 1 + (−1) (3 ) + (−21)(−21−1) (3 )2 + ⋯

(1 + 3x) 2 2 2!

= 1 − 3 + 27 2 + ⋯

28

1 −1

(b) (4 − x) 2 (1 + 3x) 2

= (2 − − 2 + ⋯ ) (1 − 3 + 27 2 + ⋯ )
4 64 28

= 2 − 3 + 27 2 − + 3 2 − 1 2 + ⋯
4 4 8 64

= 2 − 13 + 455 2 + ⋯

4 64

Chapter 3 : Sequences And Series Mathematics SM015

1

For (4 − x) 2 , |− | < 1 → −1 < < 1 → −4 < < 4
→ − 1 < < 1
4 4
33
−1

For (1 + 3x) 2 , |3 | < 1 → −1 < 3 < 1

-1 −1 1 1
3
3

The expansion is valid: − 1 < < 1

33

Let x = 1 ,
13

1 −1 1
(4− )2
(4 − x) 2 (1 + 3x) 2 = = √ 4−
1
1+3
(1+3 )2

√ 4− = 2 − 13 + 455 2 + ⋯
1+3 4 64

√14+−3((111313)) = 2 − 13 ( 1 ) + 455 ( 1 2 + ⋯

4 13 64 13 )

51

√ 13 = 1.792067303
16

13

√51
4 = 1.792067308

√51 = 7.168 ( 4 s.f.)

Chapter 3 : Sequences And Series Mathematics SM015

Q14: 2009/2010

The sum Sn of the first n terms of an arithmetic progression is given by Sn = pn + qn2 .

The sum of the first five and ten terms are 40 and 155 respectively.
(a) Find the values of p and q.
(b) Hence, find the nth term of arithmetic progression and the values of the first
term, a and the common difference, d.

(a)

Sn = pn + qn2 S10 = 10 p +100q = 155 ...(2)
S5 = 5 p + 25q = 40 ....(1) ....(3)
(1)  2 10 p + 50q = 80
(2) − (3) : 50q = 75 p=1
2
q = 3,
2

(b)

Tn = Sn − Sn−1

=  1 n + 3 n2  −  1 (n −1) + 3 (n −1)2 
2 2  2 2

= 3n −1

a = T1 = S1 = 1 (1)+ 3 (1)2 = 2

2 2

T2 = S2 − S1 =  1 (2) + 3 (2)2  −2 = 5
2 
2

d = T2 − T1 = 3

Chapter 3 : Sequences And Series Mathematics SM015

Q15: 2010/2011

(a) Given that 1 = 0.015151515.. = p + q + s + ..., where p, q and s are the first three terms of
u

geometric progression. If p = 0.015, state the value of q and s in decimal form. Hence, find
the value of u.

1

(b) Find the expansion for 1 − x 3 up to the term x2 . State the range of x for which the

 16 

1

expansion is valid. Show that 3 8 − x = 21 − x 3 . Hence, by substituting x = 2 ,
2  16 

approximate 3 7 correct to four significant figures.

(a)
1 = 0.015151515.. = p + q + s + ...,
u

0.015 + 0.00015 + 0.0000015 + ⋯ =

Given : p = 0.015
q = 0.00015
s = 0.0000015

0.00015
= 0.015 , = 0.015 = 0.01

1 0.015
= 1 − 0.01

= 1−0.01 = 990 = 66
0.015 15

Chapter 3 : Sequences And Series Mathematics SM015

(b) (1 − 1 = 1 + (1) (− ) + (13)(13−1) (− )2 +⋯

)3
16 3 16 2! 16

= 1 − − 2 + ⋯

48 2304

The expansion is valid when

−
|16 | < 1


−1 < 16 < 1

−16 < < 16

3√8 − = 3√8 (1 − )

2 16

1

= (8 (1 − ))3

16

= 1 1

83 (1 − )3
16

1

= 2 (1 − )3

16

When = 2,

1
2 23
3 − 2 = 2 (1 − 16)

√8

3√7 = 2 (1 − 2 − 22 + ⋯ )
48 2304

= 2(0.956597222 … )

= 1.913 (4.s.f)

Chapter 3 : Sequences And Series Mathematics SM015

Q16: 2010/2011
The ninth term and the sum of first fifteen terms of an arithmetic progression are 24 and
330 respectively. Find the first term, a and the common difference, d. Hence, find the least
possible value n, such that the sum of the first n terms is greater than 500.

T9 = a + 8d = 24 (1)

S15 = 15 (2a +14d ) = 15a + 105d = 330 (2)
2

solve : d = 2 and a = 8

n (16 + (n −1)(2))  500

2
n2 + 7n − 500  0

let n2 + 7n − 500 = 0

n= −7 49 − 4(1)(− 500)
2(1)

so, n = 19.13 or n = −26.13

−26.13 19.13

< −26.13 @ > 19.13

 the least possiblevalue of n is 20

Q17: 2011/2012

(a) Given that the sum of the first n terms, Sn of a series as Sn = 1 −  1 n . Find an
3

expression for the n term. Show that the series is a geometric series and find the sum to

infinity, S .

Chapter 3 : Sequences And Series Mathematics SM015

1

(b) Expand 1 + 2  2 in the descending power of x up to the term in x−3 . Hence, by

 x

substituting x = 3 , evaluate 5
correct to three decimal places.
3

(a) Tn = Sn − Sn−1

1  1  n  1  1  n−1 
 3    3  
= − − −

=  1 n−1 −  1 n
3 3

 1  n −1 1  1 1 
 3   3  
= −

= 2  1 n−1
33

Tn 2  1 n−1 1
Tn−1 33 3
r = = 2  1 n−2 =

33

∴ The sequence is a geometric sequence with common ratio = 1
3

∴ ∞ = 1 − (1)∞ = 1 − 0 = 1

3

(b)

1
1 + 2  2 = 1 + 1  2  + 1  1  1 −1 2 2 + 1  1  1 −1 1 − 2 2 3 + 
 x 2  x  2! 2  2  x  2! 2  2  2  x 

=1+ 1 − 1 + 1 +
x 2x2 2x3

By subs. x = 3,

1

1 + 2  2 =1+ 1 − 1 + 1 +
 3 3
2(3)2 2(3)3

5 = 1.296
3

Chapter 3 : Sequences And Series Mathematics SM015

Q18: 2012/2013

Given k + 2, k − 4, k − 7 are the first three terms of geometric series. Determine the value of
k. Hence, find the sum to infinity of the series.

− 4 − 7
= + 2 = − 4
( − 4)2 = ( − 7)( + 2)
2 − 2 + 16 = 2 − 5 − 14

3 = 30

= 10

The first three terms of G.S are 12, 6 , 3, …


∞ = 1 −

61
= 12, = 12 = 2

∴ ∞ = 12 = 24

1 − 1
2

Q19: 2012/2013
The first four terms of a binomial expansion (1+ ax)n is 1+ x − 1 x2 + px3 + .... Find

2
(a) The values of a and n where n  0 .

(b) The values of p. Hence, by substituting x = 1 , show that 3
is approximately equal
42

157
to .

128

Given ( + ) = + − + +. ..



(1 + ) = 1 + + ( −1) 2 2 + ( −1)( −2) 3 3+. ..

2! 3!

Comparing

Chapter 3 : Sequences And Series Mathematics SM015

: = 1

= 1 -------- (1)



2 ∶ ( −1) 2 = −1

22

( − 1) (1)2 = −1



2− = −1
2

2 − = − 2

2 2 − = 0

(2 − 1) = 0

≠ 0 @ = 1

2

From Eq(1) : = 1 = 2

1

2

(b) p = n(n − 1) (n − 2) a3 = 1  1 − 1  1 − 2   8  = 1
2  2  2  6  2
3!

substitute x = 1 int o both sides
4

1 2 3
 
 + 2  1 2 1+ 1 − 1  1 + 11
1 4   4 2  4 2  4
 

3 1+ 1 − 1 + 1
2 4 32 128

 157
128

Q20 : 2013/2014

Chapter 3 : Sequences And Series Mathematics SM015

(a) In an arithmetic progression, the sum of first four terms is 46 and the seventh term
exceeds twice of the second term by 5. Obtain the first term and the common difference for
the progression. Hence, calculate the sum of the first ten even terms of the progression.

(b) A ball is dropped from a height of 2 m. Each time the ball hits the floor, it bounces
3

vertically to a height that is of its previous height.
4

(i) Find the height of the ball at tenth bounce.
(ii) Find the total distance that the ball will travel before eleventh bounce.

(a) 4 = 4 [2 + 3 ] = 46
2

2 + 3 = 23 − − − − − −(1)

7 − 2 2 = 5
( + 6 ) − 2( + ) = 5

= 4 − 5 − − − − − −(2)

Solve Eq(1) and Eq(2) : = 4(3) − 5 = 7
2(4 − 5) + 3 = 23
11 = 33
= 3
From Eq(2) ,

1 = 7 , 2 = 10 , 3 = 13 , 4 = 16

∴ = 7 , = 16 − 10 = 6

10
10 = 2 [2(10) + 9(6)] = 370

(b) (i) 1st bounce 1 = 2 (3)

4

2nd bounce 2 = 2 (3) (3) = 2 (3)2

4 4 4

3rd bounce 3 = 2(3)2 (3) = 2 (3)3

4 4 4

Chapter 3 : Sequences And Series Mathematics SM015

:

:

10 = 2 (3)10 = 0.1126

4

(b)

1st 2nd 3rd …… 10th

Total Distance = 2 + 2 [2 (3)] + 2 [2 (3)2] + ⋯ + 2 [2 (3)10]

4 44

= 2 + 2 10

6 (1 − (34)10)
4
= 2+2
[ 1 − 3
4
]

= 13 3242

Q21 : 2014/2015
Using algebraic method, find the least value of n for which the sum of the first n terms of a
geometric series

0.88 + (0.88)2 + (0.88)3 + (0.88)4 + ...
is greater than half of its sum to infinity.

Given 0.88 + (0.88)2 + (0.88)3 + (0.88)4 + ...

= 0.88 0.882
= 0.88 = 0.88

> 1 ∞
2

Chapter 3 : Sequences And Series Mathematics SM015

(1− ) > 1 ( )

1− 2 1−

1− > 1

2

1 − 0.88 > 1
2

0.88 < 1

2

1
log 0.88 < log 2

> log21

log 0.88

> 5.422

= 6

Q22 : 2014/2015

3

(a) State the interval for x such that the expansion for (4 + 3x) 2 is valid.

3

(b) Expand (4 + 3x)2 in ascending power of x up to the term in x3

3

(c) Hence, by substituting an appropriate value of x, evaluate (5) 2 correct to three decimal
places.

(a) (4 + 3 = [4(1 + 3 3 The expansion is valid when
⌈3 ⌉ < 1
3 )2 4 ]2
4
= 3 (1 + 3 ) 3
2 − 4 < < 4
42 4
33
3

= 8 (1 + 3 )2

4

(b) (4 + 3 = 8 (1 + 3 3

3 )2 4 )2

= 8 [1 + 3 (3 ) + (32)(32−1) (3 2 + (23)(23−1)(32−2) (3 3 + ⋯ ]

24 2! 4 ) 3! 4 )

= 8 [1 + 9 + 27 2 − 27 3 + ⋯ ]
8 128 1024

Chapter 3 : Sequences And Series Mathematics SM015

= 8 + 9 + 27 2 − 27 3 + ⋯

16 128

33

(4 + 3 )2 = 52

4 + 3 = 5

= 1

3

3
(4 + 3 (1))2 = 8 + 9 (1) + 27 (1)2 − 27 (1)3 + ⋯
3 3 16 3 128 3

3

52 = 11.1797

= 11.180

Q23: 2015/2016

The first three terms of a geometric series are  4 m − 2, (2m −1) and 12.
3 

Determine the value of m. Hence, find the sixth terms for this sequence.

Given (4 − 2) , (2 − 1) , 12

3

= 2 − 1 12
=
4 2 − 1
3 − 2

(2 − 1)2 = 12 (43 − 2)

4 2 − 4 + 1 = 16 − 24

4 2 − 20 + 25 = 0

(2 − 5)2 = 0

= 5

2

Therefore : 4 ,4 , 12
3

Chapter 3 : Sequences And Series Mathematics SM015

= 4 , = 4 = 3

3 4

3

6 = 5 = 4 (3)5 = 324
(3)

Q24 : 2015/2016

(a) Expand (2 + −1 in ascending power of x, up to the term x3

x) 2

(b) Use the expansion in (a) to approximate 2
3

−1 −1

(a) (2 + ) 2 = (2 (1 + )) 2

2

= 1 [1 + (−1) ( ) + (−21)(−21−1) ( )2 + (−21)(−21−1)(−21−2) ( )3 + ⋯ ]
√2 2 2 2! 2 3! 2

= 1 [1 − + 3 2 − 5 3 + ⋯ ]

√2 4 32 128

= 1 − + 3 2 − 5 3 + ⋯
√2 4√2 32√2 128√2

1 −1

(b) Given √2 = (2)2 = (3) 2
33 2

−1 −1

Compare (2 + ) 2 = (3) 2

2

2 + = 3

2

= 3 − 2 = −1

22

We use = −1

2

−1

Thus, √2 = (2 + (−1)) 2

32

= 1 − (−21) + 3 (−1)2 − 5 (−1)3 + ⋯ = 0.8155 ≈ 0.816
√2 4√2 32√2 2 128√2 2

Q25 : 2016/2017

Chapter 3 : Sequences And Series Mathematics SM015

The seventh term of a geometric series is 16, the fifth term is 8 and the sum of first ten

terms is positive. Find the first term and the common ratio. Hence, show that

( )S12 = 126 2 +1

7 = 6 = 16 − − − − − − − −(1)
5 = 4 = 8 − − − − − − − −(2)

(2) 6 16
(1) ∶ 4 = 8

2 = 2
= ±√2

Thus, = 8 = 2
(±√2)2

( 12 − 1)
12 = − 1

2 ((√2)12 − 1)
=

√2 − 1
= 2(64−1) × √2+1

√2−1 √2+1

= 126(√2+1)

2−1

= 126(√2 + 1) shown

Chapter 3 : Sequences And Series Mathematics SM015

Q26: 2016/2017

1

(a) Obtain the expansion for (1 − x ) 4 in ascending powers of x up to the term x3 . State the
4

1

interval for x such that the expansion (1 − x ) 4 is valid. Hence, obtain the simplest form of
4

1

the expansion (16 − 4x) 4

(b) Write 4 12 in the form of K  x ) 1  . Hence, approximate 4 12 correct to three
(1 − 4 4 



decimal places

1 = 1 + (1) (− ) + (14)(14−1) (− )2 + (14)(14−1)(41−2) (− )3 + ⋯
44 2! 4 3! 4
(a) (1 − x ) 4
4

= 1 − − 3 2 − 7 3 + ⋯
16 512 8192

The expansion is valid when |− | < 1

4

−1 < < 1

4

−4 < < 4

1

(16 − 4x) 4 = [16 (1 − 1 )]

4

1 1

= 164 (1 − 1 )4

4

1

= 2 (1 − 1 )4

4

= 2 [1 − − 3 2 − 7 3 + ⋯ ]
16 512 8192

= 2 − − 3 2 − 7 3 + ⋯
8 256 4096

Chapter 3 : Sequences And Series Mathematics SM015

1

1

(b) 4 12 = 124 = (16 − 4x) 4

16 − 4 = 12
4 = 4
= 1

4 12

= ( − (1))

1

= 2 (1 − 1 (1))4

4

= 2 [1 − (1) − 3 (1)2 − 7 (1)3 + ⋯ ]
16 512 8192

= 2(0.9307861328)

= 1.8616

= 1.862

Q27: 2017/2018

The first and three more successive terms in geometric progression are given as follows:
7, …, 189, y , 1701, …

Obtain the common ratio r. Hence, find the smallest integer n such that the n-th term
exceeds 10 000

Given 7, …, 189, y , 1701, …

= = 1701

189

2 = 321489

= 567

= 567 = 3

189

> 10000
−1 > 10000
7(3) −1 > 10000

Chapter 3 : Sequences And Series Mathematics SM015

3 −1 > 10000

7

( − 1) log 3 > log 10000

7

− 1 > log100700

log 3

− 1 > 6.61
> 7.76
= 8

Q28: 2016/2017

1

(a) Expand (1 − x ) 2 in ascending powers of x up to the term x3 and state the interval of x
3

for which the expansion is valid.

1

(b) From part (a), express 9 − 3x in the form of a (1 − x) 2 , where a is an integer.
3

(c) Hence, by substituting the suitable value of x, approximate 8.70 correct to two decimal
places

(a) 1 = 1 + (1) (− ) + (12)(21−1) (− )2 + (21)(12−1)(12−2) (− )3 + ⋯
23 2! 3 3! 3
(1 − x ) 2
3

= 1 − − 1 2 − 1 3 + ⋯
6 72 432

The expansion is valid when ⌈− ⌉ < 1

3

⌈ ⌉ < 1

3

−1 < < 1

3

−3 < < 3

(b) 9 − 3x = (9 − 1 = (9(1 − 3 1 = 1 (1 − 1 = 3 (1 − 1

3 )2 ))2 92 )2 )2
9 33

(c) √8.7 = √9 − 3

Comparing : 9 − 3 = 8.7

= 0.1

Chapter 3 : Sequences And Series Mathematics SM015

Therefore, √8.7 = √9 − 3(0.1)

1

= 3 (1 − 0.1)2

3

= 3 [1 − 1 (0.1) − 1 (0.1)2 − 1 (0.1)3 + ⋯ ]
6 72 432

= 3(0.9831921290)

= 2.949576389

= 2.95 ( 2 d.p)

Q29: 2018/2019

(a) The first three terms of a geometric series are  3c − 7  , (3c − 2) and 6. Determine the
 2 

value of c. Hence, find the seventh term of this series.

(b) Expand  3 x2 −13
 2

(a)

= 3 − 2 6
=
7 3 − 2
3 − 2

(3 − 2)2 = 6 (3 − 7)

2

9 2 − 30 + 25 = 0

(3 − 5)2 = 0

= 5

3

1st term = 3 (5) − 7 = 3

3 22

Common ratio , = 6 = 2
3(35)−2

Thus, 7 = 6 = 3 (2)6 = 96
2

Chapter 3 : Sequences And Series Mathematics SM015

(b) (3 2 − 3 = (03) (3 3 (−1)0 + (03) (3 2 (−1)1 + (03) (3 1 (−1)2 +

2 1) 2 2) 2 2) 2 2)

(03) (3 2 0 (−1)3

2 )

= 27 6 − 27 4 + 9 2 − 1

8 42

Q30 : 2018/2019

(a) Find the sum of all integers from 5 to 950 which are divisible by 3.

1

(b) Expand 3(1+ x)4 in ascending powers of x up to the fourth term. Hence, approximate

4 80 correct to four decimal places correct to four decimal places

(a) Integers divisible by 3 : 6, 9, 12, 15, …, 945, 948
= 6 = 3

= 948
+ ( − 1) = 948

6 + ( − 1)3 = 948

3 = 945

= 315

315 = 315 [2(6) + (315 − 1)3] = 150255

2

(b) 1 = 3 [1 + (1) + (14)(41−1) 2 + (41)(41−1)(41−2) 3 + ⋯ ]
4 2! 3!
3(1+ x)4

= 3 [1 + 1 − 3 2 + 7 3 + ⋯ ]
4 32 128

= 3 + 3 − 9 2 + 21 3 + ⋯
4 32 128

The expansion is valid when | | < 1 ⇒ −1< < 1

Chapter 3 : Sequences And Series Mathematics SM015

4 80 1

= 804

1

1

We compare with 3(1+ x)4 = (34(1 + ))4

1

= (81 + 81 )4

11

Compare (81 + 81 )4 = 804
81 + 81 = 80
= −1

81

1 −1 2 −1 3
(81 ) (81 )
1 = (34(1 + −1 4 = 3 + 3 −1 − 9 + 21 + ⋯ = 2.9907
(81 )) 4 (81 ) 32 128
804

Q31 : 2019/2020
The sum of the first n terms of a sequence is given by Sn = 2 + 3−4n

(a) Find the value of the constant c such that n-th term is c3−4n .
(b) Show that the sequence is a geometric series.
(c) Find the sum of the infinite series, S

(a) Given Sn = 2 + 3−4n
= − −1
= (2 + 3−4 ) − (2 + 3−4( −1))
= 3−4 − 3−4 ∙ 34
= 3−4 − 81(3−4 )
= −80 (3−4 )

Comparing with Tn = c3−4n
The value of c =−80

Chapter 3 : Sequences And Series Mathematics SM015

(b) = −80 (3−4 )

= = −80 (3−4 ) = 3−4 = 3−4 −(4−4 ) = 3−4 = 1
−1 −80 (3−4( −1)) 34−4 81

Since = 1 is a constant , therefore the sequence is Geometric Series

−1 81

(c) = −80 (3−4 )

1 = −80 (3−4(1)) = − 80
81

= 1

81

−80 −80

∞ = 81 = 81 = −1
80
1−811 81