Question

The sum of the first $n$ terms of the series $1^2+2\times 2^2+3^2+2\times 4^2+5^2+2\times 6^2+\dots$ is $\frac{n(n+1{)}^2}{2}$, when $n$ is even. When $n$ is $11,$ the sum is

Answer 1

By observation,
When $n$ is even, the last term is $2n^2$
When $n$ is odd, the last tern is $n^2$

For $n$ is even,
$S=1^2+2\times 2^2+3^2+2\times 4^2+\dots +2\times n^2\Rightarrow S=\frac{n(n+1{)}^2}{2}$

Now,
$S_{11}=S_{10}+T_{11}\Rightarrow S_{11}=\frac{10\times (11{)}^2}{2}+{11}^2\therefore S_{11}=726$