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)

$\frac{n^{2}\left(n+1\right)}{2}$

Answer 2

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