Question

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

I had been asked this question in a job interview.

Asked question is from Sum to n Terms of Special Series in chapter Sequences and Series of Mathematics – Class 11

Answer 1

Correct option is (d) 966

Explanation: General term of above series is ak = 2k*(k^2+2) = 2k^3+4k

Taking summation from k=1 to k=n on both sides, we get

\(\sum_{i=0}^na_k = 2\sum_{i=0}^nk^3 + 4\sum_{i=0}^nk = 2(\frac{n(n+1)}{2})^2 + 4\frac{n(n+1)}{2}\)

= n^2(n+1)^2/2+2n(n+1)

= 36*49/2 + 2*6*7

= 966.