Question

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

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

Answer 1

Answer: (C)

If n is odd, the required sum is
$1^2+2.2^2+3^2+2.4^2+\dots \dots +2{\left(n-1\right)}^2+n^2$
$=\frac{\left(n-1\right){\left(n-1+1\right)}^2}{2}+n^2$ ($\because n-1$ is even)
$=\left(\frac{n-1}{2}+1\right)n^2=\frac{n^2\left(n+1\right)}{2}$