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The sum of first $n$ terms of the series

$1^{2}+2.2^{2}+3^{2}+2.\: 4^{2}+5^{2}+2.\: 6^{2}+......$ is

$\frac{n\left ( n+1 \right )^{2}}{2}$ when $n$ is even. When $n$ is odd the sum is
Option: 1
$\frac{n\left ( n+1 \right )^{2}}{4}$

Option: 2
$\frac{n^{2}\left ( n+1 \right )}{2}$

Option: 3
$\frac{3n\left ( n+1 \right )}{2}$

Option: 4
$\left [ \frac{n\left ( n+1 \right )}{2} \right ]^{2}$

Answers (1)

best_answer

$1^{2}+2\times 2^{2}+3^{3}+2\times 4^{2}+5^{2}+2\times 6^{2}+...$

Let n=2m

=> $1^{2}+2\times 2^{2}+3^{2}.+2.4^{2}+...$ to 2m terms= $\frac{2m(2m+1)}{2}$

$\therefore$ Now sum of odd terms is

$S_{odd}=(1^{2}+2.2^{2}+3^{2}+2.4^{2}+...2m\:terms)+(2m+1^{2})$

$=\frac{2m(2m+1)^{2}}{2}+(2m+1)^{2}$

$\therefore \frac{(n-1)(n)^{2}}{2}+n^{2}$ $[2m+1=n, \:2m=n-1]$

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

Posted by

HARSH KANKARIA

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