Compute $\sum _{i=0}^n \left(\frac{i}{n}\right)^3=\frac{(n+1)^2}{4 n}$

Question

Can someone show me how one can deal with this get the answer provided? $$\sum _{i=0}^n \left(\frac{i}{n}\right)^3=\frac{(n+1)^2}{4 n}$$ Thanks

Answer 1

The equality is equivalent to

$$\sum_{i=1}^n i^3=\frac14 n^2(n+1)^2$$ and we can prove it by induction. For the inductive step we have

$$\sum_{i=1}^{n+1} i^3=\sum_{i=1}^n i^3+(n+1)^3=\frac14 n^2(n+1)^2+(n+1)^3=\cdots\cdots$$

Answer 2

Multiply both sides by $n^3$, then use complete induction.