Prove the formula for the sum of consecutive cubes

Question

$$\sum_{k=1}^n k^3=\frac{n^2 (n+1)^2}{4}$$

Please help

Answer 1

If $n=1$, it holds; Suppose that $n=m$, the equality also holds, i.e., $$\sum_{k=1}^mk^3=\frac{m^2(m+1)^2}{4}.$$

Now $$\sum_{k=1}^{m+1}k^3=\frac{m^2(m+1)^2}{4}+(m+1)^3=\frac{(m+1)^2(m+2)^2}{4}.$$