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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 $\frac{n(n+1)^2}{2}$ if n is even . When n is odd then the sum is

A. $\frac{3n(n+1)}{2}$ B. $\frac{n^2(n+1)}{2}$ C. $\frac{n(n+1)^2}{4}$ D. $\left [ \frac{n(n+1)}{4} \right ]^2$

Answers (1)

@Jeetu

if n is odd n-1 is even

sum of odd=

$\frac{(n-1)n^2}{200}+n^2=\frac{n^2(n+1)}{2}$

Posted by

Safeer PP

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