If the sum of the first n terms of the series | vedclass

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(A) The given series is $S_n = 1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots + T_n$. When $n$ is even,the sum is given as $S_n = \

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

Vedclass · Answer

(A) The given series is $S_n = 1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots + T_n$. When $n$ is even,the sum is given as $S_n = \frac{n(n+1)^2}{2}$. When $n$ is odd,the $(n-1)$-th term is even. Thus,$S_n = S_{n-1} + T_n$. Since $n-1$ is even,$S_{n-1} = \frac{(n-1)((n-1)+1)^2}{2} = \frac{(n-1)n^2}{2}$. The $n$-th term $T_n$ for odd $n$ is $n^2$. Therefore,$S_n = \frac{(n-1)n^2}{2} + n^2 = \frac{n^2(n-1+2)}{2} = \frac{n^2(n+1)}{2}$.

If the sum of the first $n$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \dots$ is $\frac{n(n+1)^2}{2}$ when $n$ is even,what is the sum when $n$ is odd?

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