If the sum of the series 1^2 + 2 cdot 2^2 + 3 | vedclass

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(C) For an odd $n$,the series consists of $n$ terms: $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \dots + 2 \cdot (n-1)^2 + n^2$.This can be written as th

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

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(C) For an odd $n$,the series consists of $n$ terms: $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \dots + 2 \cdot (n-1)^2 + n^2$.This can be written as the sum of the first $(n-1)$ terms plus the $n$-th term.Since $(n-1)$ is even,we use the given formula for even terms: $S_{n-1} = \frac{(n-1)(n-1+1)^2}{2} = \frac{(n-1)n^2}{2}$.Now,adding the $n$-th term $(n^2)$: $S_n = \frac{(n-1)n^2}{2} + n^2$.Factoring out $n^2$: $S_n = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.

If the sum of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots + 2 \cdot (n-1)^2 + n^2$ (when $n$ is odd) is to be determined,given that for even $n$,the sum is $\frac{n(n+1)^2}{2}$,find the sum when $n$ is odd.

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