Question

The sum of first n terms of the series $1^2+2\cdot 2^2+3^2+2\cdot 4^2+5^2+2\cdot 6^2+7^2+\cdots$ is $\frac{n{(n+1)}^2}{2}$ when $n$ is even,when $n$ is odd,the sum is

• A. $\frac{n{(n+1)}^2}{2}$
• B. $\frac{n^2(n+1)}{2}$
• C. $\frac{n^2{(n+1)}^2}{4}$
• D. $\frac{n^2{(n+1)}^2}{2}$

Answer 1

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

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

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