If the sum of the series $1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+$ ... $2.6^2+\ldots$ upto $\mathrm{n}$ terms,…
If the sum of the series $1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+$ ... $2.6^2+\ldots$ upto $\mathrm{n}$ terms, when $\mathrm{n}$ is even, is $\frac{n(n+1)^2}{2}$, then the sum of the series, when $\mathrm{n}$ is odd, is
$n^2(n+1)$
$\frac{n^2(n-1)}{2}$
$\frac{n^2(n+1)}{2}$
$n^2(n-1)$
Solution
If $n$ is odd, the required sum is
$
\begin{aligned}
& 1^2+2.2^2+3^2+2.4^2+\ldots . .+2(n-1)^2+n^2 \\
& =\frac{(n-1)(n-1+1)^2}{2}+n^2 \quad(\because n-1 \text { is even }) \\
& =\left(\frac{n-1}{2}+1\right) n^2=\frac{n^2(\mathrm{n}+1)}{2}
\end{aligned}
$