250+ TOP MCQs on Sum to n Terms of Special Series & Answers | Class 11 Maths

Mathematics Written Test Questions and Answers for Class 11 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)
Answer: a
Clarification: Sum of first n terms = 1+2+3+4+……+n
=> (n/2) (a + a_n) = (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)
Answer: c
Clarification: 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)
Answer: c
Clarification: 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 = (4sum_{i=0}^n k^3 + 6 sum_{i=0}^n k^2 + 4 sum_{i=0}^n k + n)
4n³+n⁴+6n²+4n = (4sum_{i=0}^n k^3 + 6frac{(n(n+1)(2n+1))}{6} + 4frac{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
Answer: c
Clarification: 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
Answer: d
Clarification: 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 n^th 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})
Answer: b
Clarification: Given, n^th term is n(n-2)
So, a_k = 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} – 2frac{n(n+1)}{2} = frac{n(n+1)(2n-5)}{6}).

7. Find the sum of series up to 6^th term whose n^th term is given by n² + 3ⁿ.
a) 91
b) 1284
c) 1183
d) 1092
Answer: c
Clarification: Given, n^th term is n² + 3ⁿ
So, a_k = k² + 3^k
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 6^th term = 6*7*13/6 + (3/2) (3⁶-1) = 91+1092 = 1183.

8. Find the sum up to 7^th term of series 2+3+5+8+12+………………….
a) 70
b) 490
c) 340
d) 420
Answer: a
Clarification: S_n = 2+3+5+8+12+……………………………+ a_n
S_n = 2+3+5+8+12+ ……. + a_n-1 + a_n
Subtracting we get, 0 = 2+1+2+3+4+………………………….. – a_n
=>a_n = 2+1+2+3+4+…………….+(n-1) = 2+(n-1)n/2 = (1/2) (n²-n+4)
n^th term is (1/2) (n²-n+4)
So, a_k = (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
Answer: d
Clarification: General term of above series is a_k = 2k*(k²+2) = 2k³+4k
Taking summation from k=1 to k=n on both sides, we get
(sum_{i=0}^na_k = 2sum_{i=0}^nk^3 + 4sum_{i=0}^nk = 2(frac{n(n+1)}{2})^2 + 4frac{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
Answer: b
Clarification: 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
Answer: a
Clarification: 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
Answer: b
Clarification: 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
Answer: a
Clarification: 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.

Mathematics Written Test Questions and Answers for Class 11,