Question

If n is odd then find sum of n terms of $s=1^2+{2.2}^2+3^2+{2.4}^2+.....$

Answer 1

given that $n$ is add

$S=1^2+{2.2}^2+3^2+{2.4}^2+....+n^2$

if $n$ is add than $(n-1)$ is even

$S=1^2+{2.2}^2+3^2+{2.4}^2+....+2.(n-1{)}^2+n^2$

$S=1^2+(2^2+2^2)+3^2+\left(4^2{+}^2\right)+.....\left(\right(n-1{)}^2+(n-1{)}^2)+n^2$

$S=(1^2+2^2+....+n^2)+(2^2+4^2+....+(n-1{)}^2)$

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

We know that

$1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

$S=\frac{n(n+1)(2n+1)}{6}+2^2\times \frac{\left(\frac{n-1}{2}\right)(\frac{n-1}{2}+1)(2\times \frac{n-1}{2}+1)}{6}$

$S=\frac{n(n+1)(2n+1)}{6}+2^2\times \frac{\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)n}{6}$

$S=\frac{n(n+1)}{6}[2n+1+n-1]$

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