Mathematics Questions and Answers – Sum to n Terms of Special Series

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This set of Mathematics Written Test Questions and Answers for Class 11 focuses on “Sum to n Terms of Special Series”.

1. Find the sum of first n terms.
a) \(\frac{n(n+1)}{2}\)
b) \((\frac{n(n+1)}{2})^3\)
c) \(\frac{n(n+1)(2n+1)}{6}\)
d) \((\frac{n(n+1)}{2})^2\)
View Answer

Answer: a
Explanation: Sum of first n terms = 1+2+3+4+……+n
=> (n/2) (a + an) = (n/2) (1+n) = \(\frac{n(n+1)}{2}\).
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2. Find the sum of squares of first n terms.
a) \(\frac{n(n+1)}{2}\)
b) \((\frac{n(n+1)}{2})^3\)
c) \(\frac{n(n+1)(2n+1)}{6}\)
d) \((\frac{n(n+1)}{2})^2\)
View Answer

Answer: c
Explanation: Sum of squares of first n terms = 12+22+32+……………+n2
k3–(k – 1)3=3k2–3k + 1
On substituting k = 1, 2, 3, ……, n and adding we get,
n3 = 3 \(\sum_{i=0}^n k^2 – 3 \sum_{i=0}^n k + n\)
n3 = 3 \(\sum_{i=0}^n k^2 – 3 \frac{n(n+1)}{2}\) + n
\(\sum_{i=0}^n k^2 = \frac{n(n+1)(2n+1)}{6}\).

3. Find the sum of cubes of first n terms.
a) \(\frac{n(n+1)}{2}\)
b) \((\frac{n(n+1)}{2})^3\)
c) \(\frac{n(n+1)(2n+1)}{6}\)
d) \((\frac{n(n+1)}{2})^2\)
View Answer

Answer: c
Explanation: Sum of cubes of first n terms = 13+23+33+……………+n3
(k + 1)4–k4 = 4k3 + 6k2 + 4k + 1.
On substituting k = 1, 2, 3, ……, n and adding we get,
4n3+n4+6n2+4n = \(4\sum_{i=0}^n k^3 + 6 \sum_{i=0}^n k^2 + 4 \sum_{i=0}^n k + n\)
4n3+n4+6n2+4n = \(4\sum_{i=0}^n k^3 + 6\frac{(n(n+1)(2n+1))}{6} + 4\frac{n(n+1)}{2} + n\)
\(\sum_{i=0}^n k^3 = (\frac{n(n+1)}{2})^2\).
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4. Find the sum 12+22+32+……………+102.
a) 325
b) 365
c) 385
d) 435
View Answer

Answer: c
Explanation: We know, sum of squares of first n terms is given by \(\frac{(n(n+1)(2n+1))}{6}\).
Here, n=10. So, sum = 10*11*21/6 = 385.

5. Find the sum 13+23+33+……………+83.
a) 1225
b) 1184
c) 1475
d) 1296
View Answer

Answer: d
Explanation: We know, sum of cubes of first n terms is given by \((\frac{n(n+1)}{2})^2\).
Here, n=8. So, sum = (8*9/2)2 = 1296.
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6. Find the sum to n terms of the series whose nth term is n (n-2).
a) \(\frac{n(n-1)(2n+4)}{6}\)
b) \(\frac{n(n+1)(2n-5)}{6}\)
c) \(\frac{(n-2)(2n-5)}{3}\)
d) \(\frac{n(n+1)(2n-5)}{3}\)
View Answer

Answer: b
Explanation: Given, nth term is n(n-2)
So, ak = k(k-2)
Taking summation from k=1 to k=n on both sides, we get
\(\sum_{i=0}^n a_k = \sum_{i=0}^n k^2 – 2 \sum_{i=0}^n k = \frac{n(n+1)(2n+1)}{6} – 2\frac{n(n+1)}{2} = \frac{n(n+1)(2n-5)}{6}\).

7. Find the sum of series up to 6th term whose nth term is given by n2 + 3n.
a) 91
b) 1284
c) 1183
d) 1092
View Answer

Answer: c
Explanation: Given, nth term is n2 + 3n
So, ak = k2 + 3k
Taking summation from k=1 to k=n on both sides, we get
\(\sum_{i=0}^na_k = \sum_{i=0}^nk^2 + \sum_{i=0}^n3^k\)
\(\sum_{i=0}^nk^2 = n(n+1) (2n+1)/6\)
\(\sum_{i=0}^n3^k = 3*(3^n-1)/ (3-1) = (3/2) (3^n-1)\)
So, \(\sum_{i=0}^na_k = n(n+1) (2n+1)/6 + (3/2) (3^n-1)\)
Sum up to 6th term = 6*7*13/6 + (3/2) (36-1) = 91+1092 = 1183.
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8. Find the sum up to 7th term of series 2+3+5+8+12+………………….
a) 70
b) 490
c) 340
d) 420
View Answer

Answer: a
Explanation: Sn = 2+3+5+8+12+……………………………+ an
Sn = 2+3+5+8+12+ ……. + an-1 + an
Subtracting we get, 0 = 2+1+2+3+4+………………………….. – an
=>an = 2+1+2+3+4+…………….+(n-1) = 2+(n-1)n/2 = (1/2) (n2-n+4)
nth term is (1/2) (n2-n+4)
So, ak = (1/2) (k2-k+4)
Taking summation from k=1 to k=n on both sides, we get
\(\sum_{i=0}^na_k = (1/2)\sum_{i=0}^nk^2 – (1/2)\sum_{i=0}^nk + 2n\) = n(n+1) (2n+1)/(2*6) – n(n+1)/4 + 2n
Here, n=7. So, \(\sum_{i=0}^na_k\) = (7*8*15)/12 – (7*8)/4 + 2*7 = 70.

9. Find the sum to 6 terms of each of the series 2*3+4*6+6*11+8*18+………………..
a) 784
b) 882
c) 928
d) 966
View Answer

Answer: d
Explanation: General term of above series is ak = 2k*(k2+2) = 2k3+4k
Taking summation from k=1 to k=n on both sides, we get
\(\sum_{i=0}^na_k = 2\sum_{i=0}^nk^3 + 4\sum_{i=0}^nk = 2(\frac{n(n+1)}{2})^2 + 4\frac{n(n+1)}{2}\)
= n2(n+1)2/2+2n(n+1)
= 36*49/2 + 2*6*7
= 966.
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10. Find the sum of series 62+72+…………………..+152.
a) 55
b) 1185
c) 1240
d) 1385
View Answer

Answer: b
Explanation: 62+72+………………..…..+152
= (12+22+32+……..+152) – (12+22+32+42+52)
= 15*16*31/6 – 5*6*11/6
= 1240-55 = 1185.

11. Find the sum of series 63+73+………………..…..+203.
a) 43875
b) 83775
c) 43775
d) 43975
View Answer

Answer: a
Explanation: 63+73+………………..…..+203
= (13+23+33+……..+203) – (13+23+33+43+53)
= (20*21/2)2 – (5*6/2)2
= (210)2 – (15)2
= 225*195
= 43875.

12. Find the sum of series 12+32+52+…………………………..+112.
a) 279
b) 286
c) 309
d) 409
View Answer

Answer: b
Explanation: 12+32+52+…………………………..+112
= (12+22+32+……+112) – (22+42+62+82+102)
= (12+22+32+……112) – 22(12+22+32+42+52)
= 11*12*23/6 – 4*5*6*11/6
= 506 – 220
= 286.

13. Find the sum of series 13+33+53+…………………………..+113.
a) 2556
b) 5248
c) 6589
d) 9874
View Answer

Answer: a
Explanation: 13+33+53+…………………………..+113
= (13+23+33+……+113) – (23+43+63+83+103)
= (13+23+33+……113) – 23(13+23+33+43+53)
= (11*12/2)2 – 8(5*6/2)2
= 662-8*152
= 4356 – 1800
= 2556.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter