Question

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

• A. $\frac{n{(n+1)}^2}{4}$
• B. $\frac{n^2(n+1)}{2}$
• C. $\frac{3n(n+1)}{2}$
• D. $\frac{n^2{(n+1)}^2}{4}$

Answer 1

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

As $S_n$ is needed when $n$ is odd let $n=2k+1$
$\therefore S_n=S_{2k+1}=$ sum up to $2k$ terms $+(2k+1{)}^{th}$ term
$=\frac{2k(2k+1{)}^2}{2}+$ last term
$=\frac{(n-1)n^2}{2}+n^{2}asn=2k+1$
$=\frac{n^2(n+1)}{2}$
Hence, option 'B' is correct.