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

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

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.

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