Question

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

I got this question in exam.

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

Answer 1

Correct answer is (b) $\frac{n(n+1)(2n-5)}{6}$

To explain: Given, n^th term is n(n-2)

So, ak = k(k-2)

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

$\sum_{i=0}^n a_k = \sum_{i=0}^n k^2 – 2 \sum_{i=0}^n k = \frac{n(n+1)(2n+1)}{6} – 2\frac{n(n+1)}{2} = \frac{n(n+1)(2n-5)}{6}$.