H2 Mathematics – Sequences and...

H2 Mathematics – Sequences and Series

Solutions (2007 – 2016)

Summation of Series

1. 2007/P2/2

(i) Removed from syllabus.

(ii)NP

n=1

2n+ 1

n2(n+ 1)2

= 1� 1

(N + 1)2

(iii) As N ! 1,1

(N + 1)2! 0.

Therefore, the series is convergent and the sum to infinity is 1.

(iv)NX

n=2

2n� 1

n2(n� 1)2=

n+1=NX

n+1=2

2(n+ 1)� 1

(n+ 1)2 (n+ 1� 1)2(replace n with n+ 1)

=N�1X

n=1

2n+ 1

n2(n+ 1)2

= 1� 1

N2

c�infinitymaths.sg 2016 1

2. 2009/P1/3

(i)1

n� 1� 2

n+

1

n+ 1=

n(n+ 1)� 2(n� 1)(n+ 1) + n(n� 1)

(n� 1)n(n+ 1)=

2

n3 � n

(ii)nP

r=2

1

r3 � r=

1

2

nPr=2

2

r3 � r=

1

2

nPr=2

✓1

r � 1� 2

r+

1

r + 1

◆

(iii) As n ! 1,1

n! 0 and

1

n+ 1! 0.

Therefore, the series converges.

1Pr=2

1

r3 � r=

1

4

3. 2009/P1/5 partial

nPr=1

r2 = 16n(n+ 1)(2n+ 1)

2nX

r=n+1

r2 =2nX

r=1

r2 �nX

r=1

r2

= 16 (2n)(2n+ 1)(4n+ 1)� 1

6n(n+ 1)(2n+ 1)

= 16n(2n+ 1)[2(4n+ 1)� (n+ 1)]

= 16n(2n+ 1)(7n+ 1)


4. 2010/P1/3

(i) un = Sn � Sn�1

= n(2n+ c)� (n� 1) (2(n� 1) + c)

= 2n2 + cn��2n2 + cn� 4n+ 2� c

�

= 4n� 2 + c

(ii) Removed from syllabus.

5. 2010/P2/2


(ii) (a) By Partial Fractions,1

r(r + 2)=

1

2r� 1

2(r + 2)=

1

2

✓1

r� 1

r + 2

◆

nPr=1

1

r(r + 2)=

1

2

nPr=1

✓1

r� 1

r + 2

◆

=3

4� 1

2(n+ 1)� 1

2(n+ 2)

(b) As n ! 1,1

2(n+ 1)! 0 and

1

2(n+ 2)! 0.

Therefore, the series converges.

1Pr=1

1

r(r + 2)=

3

4


6. 2011/P1/6

(i) sin�r + 1

2

�✓ � sin

�r � 1

2

�✓

=�sin r✓ cos 1

2✓ + cos r✓ sin 12✓

��

�sin r✓ cos 1

2✓ � cos r✓ sin 12✓

�

= cos r✓ sin r✓ + cos r✓ sin 12✓

= 2 cos r✓ sin 12✓

(ii)nX

r=1

cos r✓ =1

2 sin 12✓

nX

r=1

2 cos r✓ sin 12✓

=1

2 sin 12✓

nX

r=1

⇥sin

�r + 1

2

�✓ � sin

�r � 1

2

�✓⇤

=sin

�n+ 1

2

�✓ � sin 1

2✓

2 sin 12✓

(iii) Removed from syllabus.


7. 2013/P1/9


(ii) f(r)� f(r � 1) = (2r3 + 3r2 + r + 24)� [2(r � 1)3 + 3(r � 1)2 + (r � 1) + 24]

= 6r2

nPr=1

6r2 =nP

r=1

⇥f(r)� f(r � 1)

⇤

nX

r=1

r2 = 16

⇥f(n)� f(0)

⇤

= 16

�2n3 + 3n2 + n

�

= 16n

�2n2 + 3n+ 1

�

= 16n (n+ 1) (2n+ 1)

(iii)nX

r=1

f(r) =nX

r=1

(2r3 + 3r2 + r + 24)

=nX

r=1

(2r3 + r) +nX

r=1

(3r2 + 24)

=nX

r=1

r(2r2 + 1) + 3nX

r=1

r2 +nX

r=1

24

= 12n(n+ 1)(n2 + n+ 1) + 1

2n(n+ 1)(2n+ 1) + 24n


8. 2014/P1/6

(a) (i) Removed from syllabus.

(ii)nX

r=1

pr =nX

r=1

13 (7� 4r)

=1

3

"nX

r=1

7�nX

r=1

4r#

=1

3

7n� 4(4n � 1)

4� 1

�

= 73n� 4

9 (4n � 1)

(b) (i) When n ! 1,1

(n+ 1)!! 0.

Therefore, the series converges. The sum to infinity is 1.

(ii) un = Sn � Sn�1

=

✓1� 1

(n+ 1)!

◆�✓1� 1

n!

◆

=1

n!� 1

(n+ 1)!

=n+ 1

(n+ 1)!� 1

(n+ 1)!

=n

(n+ 1)!


9. 2015/P2/4

(a) Removed from syllabus.

(b) (i)2

4r2 + 8r + 3=

2

(2r + 1)(2r + 3)

=A

2r + 1+

B

2r + 3

=) 2 = A(2r + 3) +B(2r + 1)

Substituting r = � 12 =) A = 1

Substituting r = � 32 =) B = �1

) 2

4r2 + 8r + 3=

1

2r + 1� 1

2r + 3

(ii) Sn =nP

r=1

2

4r2 + 8r + 3=

nPr=1

✓1

2r + 1� 1

2r + 3

◆

(iii) As n ! 1,1

2n+ 3! 0. Hence S1 = 1

3

S1 � Sn < 10�3

=) 1

3�

✓1

3� 1

2n+ 3

◆< 10�3

=) 1

2n+ 3< 10�3

=) 2n+ 3 > 103

=) n > 498.5

Therefore the smallest value of n is 499.


10. 2016/P1/6


(ii) u1 = u0 + 13 + 1 = 4

u2 = u1 + 23 + 2 = 14

u3 = u2 + 33 + 3 = 44

(iii) By the method of di↵erences,nP

r=1(ur � ur�1) = un � u0

Also,nX

r=1

(ur � ur�1) =nX

r=1

r3 + r

=nX

r=1

r(r2 + 1)

= 14n(n+ 1)(n2 + n+ 2)

Hence, un � u0 = 14n(n+ 1)(n2 + n+ 2)

=) un = 14n(n+ 1)(n2 + n+ 2) + 2


Arithmetic and Geometric Series

1. 2007/P1/10

(i) Second term of GP = Fourth term of AP:

ar = a+ 3d =) r =a+ 3d

a

Third term of GP = Sixth term of AP:

ar2 = a+ 5d =) r2 =a+ 5d

a

3r2 � 5r + 2 = 3

✓a+ 5d

a

◆� 5

✓a+ 3d

a

◆+ 2

= 3

✓1 +

5d

a

◆� 5

✓1 +

3d

a

◆+ 2

= 3 +15d

a� 5� 15d

a+ 2

= 0

(ii) 3r2 � 5r + 2 = 0

(3r � 2)(r � 1) = 0) r = 2

3 or r = 1 (rejected as d 6= 0)

Since |r| = 23 < 1 =) the geometric series is convergent

S1 =a

1� 23

= 3a

(iii) Substituting r = 23 into ar = a+ 3d :

23a = a+ 3d =) d = �a

9

S =n

2(2a+ (n� 1)d) > 4a

n

2

h2a+ (n� 1)

⇣�a

9

⌘i> 4a

na

2

✓2� n

9+

1

9

◆> 4a

n

2

✓19

9� n

9

◆> 4 (since a > 0)

n2 � 19n+ 72 < 0

=) 5.23 < n < 13.8

) {n 2 Z : 6 n 13}


2. 2008/P1/10

(i) The context describes an arithmetic progression with first term a = 10 andcommon di↵erence d = 3.

Sn =n

2(2a+ (n� 1)d) > 2000

n

2[2(10) + 3(n� 1)] > 2000

n

2(17 + 3n) > 2000

3n2 + 17n� 4000 > 0

=) n < �39.5 or n > 33.8

) It takes 34 months to save over $2000, i.e. 1 October 2011.

(ii) (a) After 2 years her original $10 has compounded into 10 (1.02)24

Amount of interest = 10 (1.02)24 � 10 = $6.08

(b) On the last day of the 24th month:

• the 1st $10 has compounded 24 times into 10 (1.02)24

• the 2nd $10 has compounded 23 times into 10 (1.02)23

• the 3rd $10 has compounded 22 times into 10 (1.02)22

• the 24th $10 has compounded once into 10 (1.02)

Total in the account = 10 (1.02) + ...+ 10 (1.02)22 + 10 (1.02)23 + 10 (1.02)24

=10(1.02)

�1.0224 � 1

�

1.02� 1= $310.30

(c)10(1.02) (1.02n � 1)

1.02� 1> 2000

1.02n > 4.9215

n lg 1.02 > lg 4.9215

n >lg 4.9215

lg 1.02

n > 80.5

) It takes 81 complete months for the total to exceed $2000.


3. 2009/P1/8

(i) Let the length of the first bar be a and the common ratio be r.

a = 20, ar24 = 5 =) r24 = 520 = 1

4 =) r =�14

� 124

S1 =a

1� r=

20

1��14

� 124

= 356.3

Therefore, the total length of all the bars must be less than 357 cm.

(ii) L = S25 =a(1� r25)

1� r=

20h1�

�14

� 2524

i

1��14

� 124

= 272.26 ⇡ 272

Length of the 13th bar = ar12 = 20�14

� 1224 = 10

(iii) Let the length of the first bar be b.

b+ (25� 1)d = 5 =) b = 5� 24d

L =25

2

⇥2b+ (25� 1)d

⇤

=) 25

2

⇥2(5� 24d) + 24d

⇤= 272.26

=) d = �0.49086

Since d < 0, the first bar is the longest bar.

Length of the longest bar = b = 5� 24(�0.49086) = 16.8 cm


4. 2011/P1/9

(i) The context describes an arithmetic progression with first term a = 256 andcommon di↵erence d = �7.

Depth drilled on the 10th day, u10 = a+ (n� 1)d

= 256 + (10� 1)(�7)

= 193m

Depth drilled on the nth day, un = 256 + (n� 1)(�7)

= 263� 7n

263� 7n < 10 =) n > 36.1

Therefore the last day is the 37th day.

Total depth drilled, S37 =n

2

⇥2a+ (n� 1)d

⇤

=37

2[2(256) + (37� 1)(�7)]

= 4810m

(ii) The context describes an geometric progression with first term a = 256 andcommon di↵erence d = 8

9 .

Total depth drilled, Sn > 99% theoretical maximum total depth, S1

=) a(1� rn)

1� r> 0.99

a

1� r1� rn > 0.99

rn < 0.01�89

�n< 0.01

n lg 89 < lg 0.01

n >lg 0.01

lg 89

n > 39.1

Therefore it takes 40 days.


5. 2012/P2/4

(i) The context describes an arithmetic progression with first term a = 100 andcommon di↵erence d = 10.

Sn =n

2(2a+ (n� 1)d) > 5000

n

2[2(100) + 10(n� 1)] > 5000

n

2(190 + 10n) > 5000

n2 + 19n� 1000 > 0

=) n < �42.5 or n > 23.5

) It takes 24 months for the value to exceed $5000, i.e. 1 December 2002.

(ii) On the last day of the nth month:

• the 1st $100 has compounded n times into 100 (1.005)n

• the 2nd $100 has compounded (n� 1) times into 100 (1.005)n�1

• the 3rd $100 has compounded (n� 2) times into 100 (1.005)n�2

• the nth $100 has compounded once into 100 (1.005)

Total value of the account = 100 (1.005)+ ...+100 (1.005)n�2 +100 (1.005)n�1 +100 (1.005)n

Using the formula for the sum of a GP,

Sn =100(1.005) (1.005n � 1)

1.005� 1

= 20 100 (1.005n � 1)

Sn = 20 100 (1.005n � 1) > 5000

1.005n > 1.2487

n lg 1.005 > lg 1.2487

n >lg 1.2487

lg 1.005

n > 44.5

) It takes 45 months for the value to exceed $5000, i.e. September 2004.

(iii) On the last day of November 2003, n = 35.

On 2 December 2003,

S35 + 100 = 5000

100x�x35 � 1

�

x� 1+ 100 = 5000

x�x35 � 1

�

x� 1= 49

From GC, x = 1.018. Therefore the required interest rate is 1.8%.


6. 2013/P1/7

(i) The context describes a geometric progression with first term a = 128 andcommon ratio d = 2

3 .

p = arn�1 = 128�23

�n�1=) ln p = ln

h128

�23

�n�1i

= ln 128 + (n� 1) ln 23

= 7 ln 2 + (n� 1) ln 2� (n� 1) ln 3

= (n+ 6) ln 2 + (�n+ 1) ln 3

) A = 1, B = 6, C = �1 and D = 1

(ii) S1 =a

1� r=

128

1� 23

= 384

Therefore, the total length of string cut o↵ can never be greater than 384 cm.

(iii) Sn =a(1� rn)

1� r=

128⇥1�

�23

�n⇤

1� 23

> 380

=) 1��23

�n> 380

384�23

�n< 1

96

n lg 23 < lg 1

96

n >lg 1

96

lg 23

= 11.3

Therefore, 12 pieces must be cut o↵ before the total length cut o↵ is greater than 380 cm.


7. 2014/P2/3

(i) (a) The context describes an arithmetic progression with first term a = 2⇥ 4 = 8 andcommon di↵erence d = 2⇥ 4 = 8.

S10 = 102 [2(8) + (10� 1)(8)] = 440

(b) Sn =n

2[2(8) + (n� 1)(8)] = 4n2 + 4n

Sn � 5000 =) 4n2 + 4n � 5000

4n2 + 4n� 5000 � 0

) n � 34.9 or n �35.9

Hence the athlete needs to complete 35 stages to run at least 5 km.

(ii) The context describes a geometric progression with first term a = 2⇥ 4 = 8 andcommon ratio r = 2.

Sn =8 (2n � 1)

2� 1= 8 (2n � 1)

Sn = 10 000 =) 8 (2n � 1) = 10 000

2n = 1251

n lg 2 = lg 1251

n = 10.3

The athlete has completed 10 stages and is running the 11th stage.

S10 =8�210 � 1

�

2� 1= 8184

10 000� 8184 = 1816

u11 = 8�210

�= 8192

8192÷ 2 = 4096

The athlete is 1816m into the 11th stage. Since he has run less than half of the 11th stage,he is at a distance 1816m from O, running away from O.


8. 2015/P1/8

(i) The context describes an arithmetic progression with first term T andcommon di↵erence d = 2.

5400 S50 6300 =) 5400 50

2[2T + (50� 1)(2)] 6300

=) 5400 50T + 2450 6300

=) 59 T 77

(ii) The context describes a geometric progression with first term t andcommon ratio r = 1.02.

5400 S50 6300 =) 5400 t⇥(1.02)50 � 1

⇤

1.02� 1 6300

=) 63.845 t 74.486

=) 63.9 t 74.4

(iii) Each athlete completes the 20 km run in exactly 1 12 hours

=) T = 59 and t = 63.9

Athlete A’s time for 50th lap = T + (50� 1)d = 59 + (49)(2) = 157 s

Athlete B ’s time for 50th lap = tr50�1 = (63.845)(1.02)49 = 168.47 s

Di↵erence = 168.47� 157 = 11.47 ⇡ 11 s


9. 2016/P1/4

(i) br4 = a+ 3d (1)

br7 = a+ 8d (2)

br14 = a+ 11d (3)

(3)

(1):br14

br4=

a+ 11d

a+ 3d=) r10 =

a+ 11d

a+ 3d

(2)

(1):br7

br4=

a+ 8d

a+ 3d=) r3 =

a+ 8d

a+ 3d

5r10 � 8r3 + 3 = 5

✓a+ 11d

a+ 3d

◆� 8

✓a+ 8d

a+ 3d

◆+ 3

=5(a+ 11d)� 8(a+ 8d)

a+ 3d+ 3

=�3a� 9d

a+ 3d+ 3

=�3(a+ 3d)

a+ 3d+ 3

= �3 + 3

= 0

Solving 5r10 � 8r3 + 3 = 0 with GC,r = 0.74 or r = 1 (rejected)

(ii) The required sum is the sum to infinity of a GP with first term brn and common ratio r = 0.74.

S1 =brn

1� r

=b(0.74n)

1� 0.74

=50b(0.74n)

13


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