Question #d2197

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Answer 1 · 2017-03-31T09:36:25

Given that the sum of first n terms of

the series is $1^2+2.2^2+3^2+2.4^2+5^2+...$ is $n(n+1)^2 /2$ when n is even

So when n is even

$S_n=1^2+2.2^2+3^2+2.4^2+5^2+...+(n-1)^2+ 2 n^2 =n(n+1)^2 /2.....[1]$

when n is odd the n th term or last term is $n^2$

Now when n is odd $n-1$ will be even. So putting $n-1$ in place of n in relation [1] we get the sum up to $n-1$ terms

Hence

$S_(n-1)=1^2+2.2^2+3^2+2.4^2+5^2+...+(n-2)^2+ 2 (n-1)^2 =(n-1)(n-1+1)^2 /2$

$=((n-1)n^2) /2$

So when n is odd the sum up to n th term will be

$S_n=S_(n-1)+n^2=((n-1)n^2) /2+n^2=n^2((n-1)/2+1)$

$=(n^2(n+1))/2$