Question

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

• A. $\frac{n^2(n+1)}{2}$
• B. $\frac{n(n^2+1)}{2}$
• C. $2{(n+1)}^2.(2n+1)$
• D. none of these

Answer 1

The correct option is A $\frac{n^2(n+1)}{2}$

Given that, $1^2+2\cdot 2^2+3^2+2\cdot 4^2+5^2+2\cdot 6^2+7^2+\cdots =\frac{n{(n+1)}^2}{2}$

where. $n$ is even

If $n$ is odd then $(n-1)$ is even and last term will be $n^2$

$\therefore$ Sum of $(n-1)$ terms $=\frac{n^2(n-1)}{2}$ (given) the $n^{th}$ term will be $n^2$

$\therefore$ Required sum$=$ sum of $(n-1)$ term$+$ last term

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