The sum of the first n terms of the series 1^ | vedclass

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(B) The given series is $1^2, 2 \cdot 2^2, 3^2, 2 \cdot 4^2, 5^2, 2 \cdot 6^2, \dots$When $n$ is even,the sum $S_n = \frac{n(n + 1)^2}{2}$.When $n$ is

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

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(B) The given series is $1^2, 2 \cdot 2^2, 3^2, 2 \cdot 4^2, 5^2, 2 \cdot 6^2, \dots$When $n$ is even,the sum $S_n = \frac{n(n + 1)^2}{2}$.When $n$ is odd,the $n^{th}$ term is $n^2$. The sum of the first $n$ terms is $S_n = S_{n-1} + a_n$.Since $n$ is odd,$n-1$ is even. Using the given formula for $n-1$ terms:$S_{n-1} = \frac{(n-1)((n-1) + 1)^2}{2} = \frac{(n-1)n^2}{2}$.Thus,$S_n = \frac{(n-1)n^2}{2} + n^2 = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.Verification: For $n=1$,$S_1 = 1^2 = 1$. Using formula $(b)$: $\frac{1^2(1+1)}{2} = 1$. Correct.

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. When $n$ is odd,the sum is:

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