Problem with infinite series sum I'm trying to calculate the infinite sum of the series 
$$\sum_{i=1}^{n}\frac{(i-1)^2}{n^3}. $$ 
Computing it with WolframAlpha results in $$\frac{2n^2-3n+1}{6n^2}.$$
I'm trying to solve it by considering it as an arithmetic series and using the formula for the nth $$\sum_{i=1}^{n}a_i=\frac{n(a_1+a_n)}{2},$$ but I can't reach the correct result. 
Any clue, bibliography, anything? 
EDITED
Following the instructions of avz2611: 
$$\frac{1}{n^3}\sum_{i=1}^{n}(i-1)^2=\frac{1}{n^3}\cdot\frac{n(n-1)^2}{2}=\frac{n^2-2n+1}{2n^2}$$
with $a_1=0$ and $a_n=(n-1)^2$.

$$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\\0^2+1^2+2^2+...+(n-1)^2=\frac{(n-1)(n-1+1)(2(n-1)+1)}{6}=\\\frac{n(n-1)(2n-1)}{6}$$so  $$ \sum_{i=1}^{n}\frac{(i-1)^2}{n^3}=\frac{1}{n^3}\sum_{i=1}^{n}(i-1)^2=\\\frac{\frac{n(n-1)(2n-1)}{6}}{n^3}=\\\frac{(n-1)(2n-1)}{6n^2}$$
