Question

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

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

Answer 1

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

The sum of $n$ terms of given series $=\frac{n{(n+1)}^2}{2}$
Let $n$ is odd, i.e., $n=2m+1$
Then, $S_{2m+1}\equiv S_{2m}+(2m+1)$th term
$=\frac{(n-1)n^2}{2}+n$th term
$=\frac{(n-1)n^2}{2}+n^2$
$=\frac{(n+1)n^2}{2}$