Question

Question #d2197

Answer 1

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`