Answer
$\Sigma^{n+1}_{i=0}\frac{(i-1)^2}{i.2}$ =$\Sigma^{n}_{j=1}\frac{j^2}{jn+n}$
Work Step by Step
When i = 1, then j = 0.
When i = n + 1, then j = n.
Since j = i − 1, then i = j + 1.
Thus,
$\frac{(i-1)^2}{i.n}=\frac{((j+1)-1)^2}{(j+1).n}=\frac{j^2}{jn+n}$
N is constant so
$\Sigma^{n+1}_{i=0}\frac{(i-1)^2}{i.2}$ =$\Sigma^{n}_{j=1}\frac{j^2}{jn+n}$