Question

The sum of the first $n$ terms of the series $1^2+2\times 2^2+3^2+2\times 4^2+5^2\times 6^2+....$ is $n(n+1{)}^2/2$, when $n$ is even. When $n$ is odd, the sum is __________.

Answer 1

When $n$ is odd, last term is $n^2$. Hence, the required sum is

$S=[1^2+2\times 2^2+3^2+2\times 4^2+...+2\times (n-1{)}^2]+n^2$

$=\frac{(n-1)n^2}{2}+n^2$ [Using sum for $(n-1)$ to be even]

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