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

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(B) The series is $1^2, 2.2^2, 3^2, 2.4^2, 5^2, 2.6^2, \dots$When $n$ is odd,the $n^{th}$ term is $n^2$.Since $n$ is odd,$n-1$ is even.The sum of the

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

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(B) The series is $1^2, 2.2^2, 3^2, 2.4^2, 5^2, 2.6^2, \dots$When $n$ is odd,the $n^{th}$ term is $n^2$.Since $n$ is odd,$n-1$ is even.The sum of the first $n-1$ terms is obtained by substituting $n-1$ for $n$ in the given formula for even terms:$S_{n-1} = \frac{(n-1)((n-1) + 1)^2}{2} = \frac{(n-1)n^2}{2}$.The sum of $n$ terms is $S_n = S_{n-1} + T_n$.$S_n = \frac{(n-1)n^2}{2} + n^2 = n^2 \left( \frac{n-1}{2} + 1 ight) = n^2 \left( \frac{n-1+2}{2} ight) = \frac{n^2(n+1)}{2}$.Verification: For $n=1$,$S_1 = 1^2 = 1$. Option $(b)$ gives $\frac{1^2(1+1)}{2} = 1$. For $n=3$,$S_3 = 1^2 + 2.2^2 + 3^2 = 1 + 8 + 9 = 18$. Option $(b)$ gives $\frac{3^2(3+1)}{2} = \frac{9 	imes 4}{2} = 18$.

The sum of the first $n$ terms of the series $1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots$ is $\frac{n(n + 1)^2}{2}$ when $n$ is even. When $n$ is odd,the sum will be:

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