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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
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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)
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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
(i) Removed from syllabus.
(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
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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.
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7. 2013/P1/9
(i) Removed from syllabus.
(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
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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)!
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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.
c�infinitymaths.sg 2016 7
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10. 2016/P1/6
(i) Removed from syllabus.
(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
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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}
c�infinitymaths.sg 2016 9
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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.
c�infinitymaths.sg 2016 10
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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
c�infinitymaths.sg 2016 11
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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.
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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%.
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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.
c�infinitymaths.sg 2016 14
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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.
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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
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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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