mt chapter 02

LEE KIAN KEONG STPM MATHEMATICS (T) 2: Sequences and Series

2 Sequences and Series

1. [STPM ]

Evaluate

∞∑r=1

1

102r. [2 marks]

Express 0.1̄8 as a rational number in its lowest form. [2 marks]

[Answer :1

99,

2

11]

2. [STPM ]

Find the expansion of(1 + x2)p

(1− x)qin ascending powers of x until the term in x3. [5 marks]

(a) If p = q =1

2, suggest a suitable value of x that enables

√13

10to be estimated using the above expansion.

Hence, estimate

√13

10correct to four decimal places. [7 marks]

(b) If p = −1

3and q lies in the interval [0,9], and the largest possible coefficient of x3. [3 marks]

[Answer : 1 + qx +

(p +

q(q + 1)

2

)x2 +

(pq +

q(q + 1)(q + 2)

6

)x3 + . . . ; (a) x =

1

5, 1.1395 ; (b) 162]

3. [STPM ]

Express (√

2− 1)5 in the form a√

2 + b, where a, b are integers. [3 marks]

[Answer : 29√

2− 41]

4. [STPM ]The sum of the first 2n terms of a series P is 20n− 4n2. Find in terms of n, the sum of the first n terms of thisseries. Show that this series is an arithmetic series. [4 marks]

Series Q is an arithmetic series such that the sum of its first n even terms is more than the sum of its first n oddterms by 4n. Find the common difference of the series Q. [5 marks]

If the first term of series Q is 1, determine the minimum value of n such that the difference between the sum ofthe first n terms of series P and the sum of the first n terms of series Q is more than 980. [6 marks]

[Answer : Sn = 10n− n2 ; d = 4 ; n = 21]

5. [STPM ]

Simplify (1 + 2√

3)5 − (1− 2√

3)5. [4 marks]

[Answer : 1076√

3]

6. [STPM ]

(a) Sn denotes the sum of the first n terms of a geometric series 3 − 1 +1

3− . . . and S∞ denotes the sum to

infinity of this series. Find the smallest n such that |S∞ − Sn| < 0.0001. [7 marks]

(b) The first and the second term of an arithmetic series equal respectively the first and the second term of ageometric series. The third term of the geometric series exceeds the third term of the arithmetic series by3. The arithmetic series has a positive common difference and the sum of its first three terms equals 54.Find the first term of both series. Find also the common difference of the arithmetic series and the commonratio of the geometric series. [8 marks]

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[Answer : (a) 10 ; (b) a = 12 , d = 6 , r =3

2]

7. [STPM ]Given that the sum of the first n terms of a series is n log pqn+1. Show that

(a) the n-th term of the series is log pq2n, [2 marks]

(b) this is an arithmetic series. [3 marks]

8. [STPM ]

Expand (1 + x)15 in ascending power of x until the term in x3. By taking x =

1

40, find the approximation for

32.815 correct to four decimal places. [7 marks]

If the expansion of1 + ax

1 + bxand (1 + x)

15 are the same until the term in x2, find the values of a and b. Hence,

show that 32.8 ≈ 203

101. [8 marks]

[Answer : 1 +1

5x− 2

25x2 +

6

125x3 + . . . , 2.0101 ; a =

3

5, b =

2

5]

9. [STPM ]

Expand (1 + 8x)12 in the ascending power of x until the term in x3. By taking x =

1

100, find

√3 correct to five

decimal places. [4 marks]

[Answer : 1 + 4x− 8x2 + 32x3 + . . . ; 1.73205]

10. [STPM ]

Given that Sn = a + ar + ar2 + . . . + arn−1, with a 6= 0. Show that Sn =a(1− rn)

1− r.

Give the condition on r such that limn→∞

Sn exists, and express this limit in terms of a and r. [5 marks]

(a) Determine the smallest integer n such that

1 +4

3+

(4

3

)2

+ . . . +

(4

3

)n

> 21.

[5 marks]

(b) Find the sum to infinity32(1− x)2 + 33(1− x)3 + . . . + 3r(1− x)r + . . .

and determine the set of x such that this sum exists. [5 marks]

[Answer : |r| < 1,a

1− r; (a) 7 ; (b)

9(1− x)2

3x− 2, {x :

2

3< x <

4

3}]

11. [STPM ]The sum and product of three consecutive terms of an arithmetic progression are −3 and 24 respectively. De-termine the three possible terms of the arithmetic progression. [5 marks]

[Answer : 4,-1,-6 or -6,-1,4]

12. [STPM ]

Expand(

1− x

n

)nwhere n is a positive integer in ascending powers of x until the term in x3. If the coefficient

of x3 is − 1

27, find n. [6 marks]

With this value of n, obtain the expansion of(

1− x

n

)n(1− x)

12 in ascending powers of x until the term in x3.

[5 marks]

Hence, by taking x = − 1

10, find the approximation of

√10 accurate to 3 decimal places. [4 marks]

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[Answer : 1− x +n− 1

2nx2 − (n− 1)(n− 2)

6n2x3 + . . . , n = 3 ; 1− 3

2x +

17

24x2 − 61

432x3 . . . ;

√10 ≈ 3.162]

13. [STPM ]

Express1

(4r − 3)(4r + 1)in partial fractions. Hence show that

n∑r=1

1

(4r − 3)(4r + 1)=

1

4

(1− 1

4n + 1

).

[6 marks]

[Answer :1

4(4r − 3)− 1

4(4r + 1)]

14. [STPM ]

Given that y =1√

1 + 2x +√

1 + x, where x > −1

2, show that, provided x 6= 0,

y =1

x(√

1 + 2x−√

1 + x).

[3 marks]

Using the second form for y, express y as a series of ascending powers of x as far as the term in x2. [6 marks]

Hence, by putting x =1

100, show that

10√102 +

√101≈ 79407

160000.

[3 marks]

[Answer : y =1

2− 3

8x +

7

16x2 + . . .]

15. [STPM ]Determine the set of x such that the geometric series 1 + ex + e2x + . . . converges. Find the exact value of x sothat the series converges to 2. [6 marks]

[Answer : {x : x < 0} ; x = − ln 2]

16. [STPM ]

Express1

4k2 − 1as partial fraction. [4 marks]

Hence, find a simple expression for Sn =

n∑k=1

1

4k2 − 1and find lim

n→∞Sn. [4 marks]

[Answer :1

2(2k − 1)− 1

2(2k + 1); Sn =

1

2

(1− 1

2n + 1

);

1

2]

17. [STPM ]

Express

(1 + x

1 + 2x

) 12

as a series of ascending powers of x up to the term in x3. [6 marks]

By taking x =1

30, find

√62 correct to four decimal places. [3 marks]

[Answer : 1− 1

2x +

7

8x2 − 25

16x3 + . . . ;

√62 = 7.8740]

18. [STPM ]

Express ur =2

r2 + 2rin partial fractions. [3 marks]

Using the result obtained,

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(a) show that u2r = −1

r+

1

r2+

1

r + 2+

1

(r + 2)2, [2 marks]

(b) show that

n∑r=1

ur =3

2− 1

n + 1− 1

n + 2and determine the values of

∞∑r=1

ur and

∞∑r=1

(ur+1 +

1

3r

). [9 marks]

[Answer :1

r− 1

r + 2; (b)

3

2,

4

3]

19. [STPM ]

Expand (1 − x)12 in ascending powers of x up to the term in x3. Hence, find the value of

√7 correct to five

decimal places. [5 marks]

[Answer : 1− 1

2x− 1

8x2 − 1

16x3 ;√

7 ≈ 2.64609]

20. [STPM ]Prove that the sum of the first n terms of a geometric series a + ar + ar2 + . . . is

a(1− rn)

1− r.

[3 marks]

(a) The sum of the first five terms of a geometric series is 33 and the sum of the first ten terms of the geometricseries is -1023. Find the common ratio and the first term of the geometric series. [5 marks]

(b) The sum of the first n terms and the sum to infinity of the geometric series 6− 3 +3

2− . . . are Sn and S∞

respectively. Determine the smallest value of n such that |Sn − S∞| < 0.001. [7 marks]

[Answer : (a) r = −2, a = 3 ; (b) n = 12]

21. [STPM ]For the geometric series 7 + 3.5 + 1.75 + 0.875 + ..., find the smallest value of n for which the different betweenthe sum of the first n terms and the sum to infinity is less than 0.01. [6 marks]

[Answer : 11]

22. [STPM ]

Express f(x) =x2 − x− 1

(x + 2)(x− 3)in partial fractions. [5 marks]

Hence, obtain an expansion of f(x) in ascending powers of1

xup to the term in

1

x3. [6 marks]

Determine the set of values of x for which this expansion is valid. [2 marks]

[Answer : 1 +1

x− 3− 1

x + 2, 1 +

5

x2+

5

x3+ . . . , {x : x < −3, x > 3}]

23. [STPM ]If x is so small that x2 and higher powers of x may be neglected, show that

(1− x)6(

2 +x

2

)10≈ 29(2− 7x).

[4 marks]

24. [STPM ]

The nth term of an arithmetic progression is Tn, show that Un =5

2(−2)2(

10−Tn17 ) is the nth term of a geometric

progression. [4 marks]

If Tn =1

2(17n− 14), evaluate

∞∑n=1

Un. [4 marks]

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[Answer : −10

3]

25. [STPM ]Express the infinite recurring decimal 0.72̇5̇ (= 0.7252525 . . . ) as a fraction in its lowest terms. [4 marks]

[Answer :359

495]

26. [STPM ]At the beginning of this year, Mr. Liu and Miss Dora deposited RM10 000 and RM2000 respectively in abank. They receive an interest of 4% per annum. Mr Liu does not make any additional deposit nor withdrawal,whereas, Miss Dora continues to deposit RM2000 at the beginning of each of the subsequent years without anywithdrawal.

(a) Calculate the total savings of Mr. Liu at the end of n-th year. [3 marks]

(b) Calculate the total savings of Miss Dora at the end of n-th year. [7 marks]

(c) Determine in which year the total savings of Miss Dora exceeds the total savings of Mr. Liu. [5 marks]

[Answer : (a) 10000(1.04)n; (b) 52000[1.04n − 1]; (c) 6]

27. [STPM ]

For geometric series 6 + 3 +3

2+ . . ., obtain the smallest value of n if the difference between the sum of the first

n + 4 terms and the sum of first n terms is less than45

64. [6 marks]

[Answer : 5]

28. [STPM ]Determine the set of values of x such that the geometric series e−x + e−2x + e−3x + . . . converges.

Find the exact value of x if the sum to infinity of the series is 3. [6 marks]

[Answer : {x : x > 0} ; x = ln4

3]

29. [STPM ]

Given f(x) =x3 − 3x− 4

(x− 1)(x2 + 1),

(a) find the constants A, B, C and D such that f(x) = A +B

x− 1+

Cx + D

x2 + 1, [5 marks]

(b) when x is sufficiently small such that x4 and higher powers can be neglected, show that f(x) ≈ 4 + 7x +3x2 − x3. [4 marks]

[Answer : (a) A = 1, B = −3, C = 4, D = 0]

30. [STPM ]

Show that

n∑r=1

r2 + r − 1

r2 + r=

n2

n + 1. [4 marks]

31. [STPM ]The sum of the first n terms of a progression 3n2. Determine the n-th term of the progression, and hence, deducethe type of progression. [4 marks]

[Answer : 6n− 3, Arithmetic Progression]

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32. [STPM ]Express in partial fractions

3

(3r − 1)(3r + 2).

[4 marks]

Show thatn∑

r=1

3

(3r − 1)(3r + 2)=

1

2− 1

(3n + 2),

[2 marks]

and hence, find∞∑r=1

1

(3r − 1)(3r + 2).

[2 marks]

[Answer :1

3r − 1− 1

3r + 2,

1

6]

33. [STPM ]

A sequence is defined by ur = e−(r−1) − e−r for all integers r ≥ 1. Find

n∑r=1

ur in terms of n, and deduce the

value of

∞∑r=1

ur. [5 marks]

[Answer : 1− e−n ; 1]

34. [STPM ]The sequence u1, u2, u3, . . . is defined by un+1 = 3un, u1 = 2.

(a) Write down the first five terms of the sequence. [2 marks]

(b) Suggest an explicit formula for ur. [2 marks]

[Answer : 2, 6, 18, 54, 162 , ur = 2(3)r−1]

35. [STPM ]

Expand (1 + x)23 and

1 + ax

1 + bx, where |b| < 1, in ascending powers of x up to the term in x3. Determine the set

of values of x for which both the expansions are valid. [7 marks]

If the two expansions are identical up to the term in x2,

(a) determine the values of a and b, [3 marks]

(b) use x =1

8to obtain the approximation

3√

81 ≈ 212

49. [3 marks]

(c) find, correct to five decimal places, the difference between the terms in x3 for the two expansions with

x =1

8. [2 marks]

[Answer : 1 +2

3x− 1

9x2 +

4

81x3 + . . . , 1 + (a− b)x + b(b− a)x2 + b2(a− b)x3 + . . ., |x| < 1 ; (a) a =

5

6, b =

1

6; (c)

0.00006]

36. [STPM ]A sequence a1, a2, a3, . . . is defined by an = 3n2. The difference between successive terms of the sequence formsa new sequence b1, b2, b3, . . ..

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(a) Express bn in terms of n. [2 marks]

(b) Show that b1, b2, b3, . . . is an arithmetic sequence, and state its first term and common difference. [3 marks]

(c) Find the sum of the first n terms of the sequence b1, b2, b3, . . . in terms of an and bn. [2 marks]

[Answer : (a) 6n + 3 , (b) 9, 6 ; (c) an + bn − 3]

37. [STPM ]Write the infinite recurring decimal 0.131̇8̇(= 0.13181818 . . .) as the sum of a constant and a geometric series.Hence, express the recurring decimal as a fraction in its lowest terms. [4 marks]

[Answer :29

220]

38. [STPM ]

(a) Express1

(r2 − 1)in partial fractions, and deduce that

1

r(r2 − 1)≡ 1

2

[1

r(r − 1)− 1

r(r + 1)

].

[4 marks]

Hence, use the method of differences to find the sum of the first (n− 1) terms, Sn−1, of the series

1

2× 3+

1

3× 8+

1

4× 15+ . . . +

1

r(r2 − 1)+ . . . ,

and deduce Sn. [6 marks]

(b) Explain why the series converges to1

4, and determine the smallest value of n such that

1

4− Sn < 0.0025.

[5 marks]

[Answer : (a)1

2(r − 1)− 1

2(r + 1); (b) Sn−1 =

1

2

(1

2− 1

n(n + 1)

), Sn =

1

2

(1

2− 1

(n + 1)(n + 2)

); (c) 13]

39. [STPM ]

Use the binomial expansions of (√

3 + 2)6 and (√

3 − 2)6 to evaluate (√

3 + 2)6 + (√

3 − 2)6. Hence, show that

2701 < (√

3 + 2)6 < 2702. [7 marks]

[Answer : 2702]

40. [STPM ]

(a) Show that for a fixed number x 6= 1, 3x2 + 3x3 + . . . + 3xn is a geometric series, and find its sum in termsof x and n. [4 marks]

(b) The series Tn(x) is given by

Tn(x) = x + 4x2 + 7x3 + . . . + (3n− 2)xn, for x 6= 1.

By considering Tn(x)− xTn(x) and using the result from (a), show that

Tn(x) =x + 2x2 − (3n + 1)xn+1 + (3n− 2)xn+2

(1− x)2.

[5 marks]

Hence, find the value of

20∑r=1

2r(3r − 2), and deduce the value of

19∑r=0

2r+2(3r + 1). [6 marks]

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[Answer : (a) sum=3x2(xn−1 − 1)

x− 1; (b) 115343370 , 230686740]

41. [STPM ]

Show that the first three terms in the expansions of (1− 8x)4 and1− 5x

1− 3xin ascending powers of x are the same.

Determine the range of values of x for which both expansions are valid. [6 marks]

Use the result (1− 8x)4 ≈ 1− 5x

1− 3xto obtain an approximation of (0.84)

14 as a fraction. [3 marks]

[Answer : {x < −1

8, x >

1

8} , 0.84

14 ≈ 45

47]

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mt chapter 02

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