Question

The sum of $n$ terms of the series $1^2+{2.2}^2+3^2+{2.4}^2+5^2+{2.6}^2+....$ is $\frac{n(n+1{)}^2}{2}$, where $n$ is even. Find the sum, where $n$ is odd ______.

Answer 1

Here, $1^2+{2.2}^2+3^2+{2.4}^2+5^2+{2.6}^2+....n^2$

$=\frac{n{(n+1)}^2}{2}$ (when $n$ is even ...(i))

When $n$ is odd

$1^2+{2.2}^2+3^2+{2.4}^2+5^2+{2.6}^2+....n^2$

$=\left\{1^2+{2.2}^2+3^2+{2.4}^2+....+2{(n-1)}^2\right\}+n^2$

$=\left\{\frac{(n-1)n^2}{2}\right\}+n^2$ [from (i)]

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

$\therefore 1^2+{2.2}^2+3^2+{2.4}^2...$ upto n terms when $n$ is odd $=n^2\frac{(n+1)}{2}$