# Question d2197

Mar 31, 2017

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 $n {\left(n + 1\right)}^{2} / 2$ 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#

$= \frac{\left(n - 1\right) {n}^{2}}{2}$

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

${S}_{n} = {S}_{n - 1} + {n}^{2} = \frac{\left(n - 1\right) {n}^{2}}{2} + {n}^{2} = {n}^{2} \left(\frac{n - 1}{2} + 1\right)$

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