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

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(C) The series is $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots$ When $n$ is odd,the sum up to $n$ terms is $S_n = (1^2 + 2 \cdot 2^2 + 3^2 + 2

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The series is $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots$ When $n$ is odd,the sum up to $n$ terms is $S_n = (1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \dots + 2 \cdot (n-1)^2) + n^2$. Since $(n-1)$ is even,we use the given formula for the sum of the first $(n-1)$ terms: $S_{n-1} = \frac{(n-1)((n-1)+1)^2}{2} = \frac{(n-1)n^2}{2}$. Therefore,the sum for $n$ terms is $S_n = \frac{(n-1)n^2}{2} + n^2$. $S_n = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.

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) or $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \dots + 2 \cdot n^2$ (when $n$ is even) is given by $S_n = \frac{n(n+1)^2}{2}$ for even $n$,find the sum of the series when $n$ is odd.

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