The sum of the first $n$ terms of the series $1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+2 \cdot 6^2+\ldots$ is…

The sum of the first $n$ terms of the series $1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+2 \cdot 6^2+\ldots$ is $\frac{n(n+1)^2}{2}$ when $n$ is even. When $n$ is odd the sum is
  1. $\frac{3 n(n+1)}{2}$
  2. $\frac{\mathrm{n}^2(\mathrm{n}+1)}{2}$
  3. $\frac{n(n+1)^2}{4}$
  4. $\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2$

Solution

If $n$ is odd then $(n-1)$ is even $\Rightarrow$ sum of odd terms $=\frac{(n-1) n^2}{2}+n^2=\frac{n^2(n+1)}{2}$.

Asked in: JEE Main 2004

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