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

«
»

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) $$(\frac{n(n+1)}{2})^2$$

Explanation: Sum of first n terms = 1+2+3+4+……+n
=> (n/2) (a + an) = (n/2) (1+n) = $$\frac{n(n+1)}{2}$$.

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) $$(\frac{n(n+1)}{2})^2$$

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) $$(\frac{n(n+1)}{2})^2$$

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 = $$4\sum_{i=0}^n k^3 + 6\frac{(n(n+1)(2n+1))}{6} + 4\frac{n(n+1)}{2} + n$$
$$\sum_{i=0}^n k^3 = (\frac{n(n+1)}{2})^2$$.

4. Find the sum 12+22+32+……………+102.
a) 325
b) 365
c) 385
d) 435

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

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.

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) $$\frac{n(n+1)(2n-5)}{3}$$

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

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.

8. Find the sum up to 7th term of series 2+3+5+8+12+………………….
a) 70
b) 490
c) 340
d) 420

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

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.

10. Find the sum of series 62+72+…………………..+152.
a) 55
b) 1185
c) 1240
d) 1385

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

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

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

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.

To practice Mathematics Written Test Questions and Answers for Class 11, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs! 