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If $S_n$ is the sum of the first $n$ terms of the series
$1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots \infty,$
then, when $n$ is even $S_n=$
Options:
$1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots \infty,$
then, when $n$ is even $S_n=$
Solution:
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Verified Answer
The correct answer is:
$\frac{n(n+1)^2}{2}$
We have,
$S_n=1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots \infty$
$\therefore T_n=(2 n-1)^2+2 \times(2 n)^2$
$=4 n^2+1-4 n+8 n^2$
$=12 n^2-4 n+1$
$\Sigma T_n=12 \Sigma n^2-4 \Sigma n+\Sigma 1$
$=12 \frac{n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n$
$=2 n(n+1)(2 n+1)-2 n(n+1)+n$
$=2 n(n+1)(2 n+1-1)+n$
$=2 n(n+1)(2 n)+n$
$=4 n^2(n+1)+n=n[4 n(n+1)+1]$
$=n\left(4 n^2+4 n+1\right)=n(2 n+1)^2$
Here, $S_n$ is even
$S_m=\frac{m}{2}\left(2 \frac{m}{2}+1\right)^2 n=\frac{m}{2} \Rightarrow S_m=\frac{m}{2}(m+1)^2$
$\therefore S_n=\frac{n}{2}(n+1)^2$ when $n$ is even
$S_n=1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots \infty$
$\therefore T_n=(2 n-1)^2+2 \times(2 n)^2$
$=4 n^2+1-4 n+8 n^2$
$=12 n^2-4 n+1$
$\Sigma T_n=12 \Sigma n^2-4 \Sigma n+\Sigma 1$
$=12 \frac{n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n$
$=2 n(n+1)(2 n+1)-2 n(n+1)+n$
$=2 n(n+1)(2 n+1-1)+n$
$=2 n(n+1)(2 n)+n$
$=4 n^2(n+1)+n=n[4 n(n+1)+1]$
$=n\left(4 n^2+4 n+1\right)=n(2 n+1)^2$
Here, $S_n$ is even
$S_m=\frac{m}{2}\left(2 \frac{m}{2}+1\right)^2 n=\frac{m}{2} \Rightarrow S_m=\frac{m}{2}(m+1)^2$
$\therefore S_n=\frac{n}{2}(n+1)^2$ when $n$ is even
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