Answer
See explanation
Work Step by Step
We shall prove that
$$\sum_{i=1}^{i=n}i^3=1^3+2^3+3^3+...+n^3=\frac{n^2(n+1)^2}{4}$$
For n=1 we have
$$1^3=\frac{1^22^2}{4}=1$$
We need to prove that
$$\sum_{i=1}^{i=n+1}=1^3+2^3+3^3+...+n^3+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$$
or
$$\sum_{i=1}^{i=n}i^3+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$$
But from induction supposition we have,
$$\sum_{i=1}^{i=n}i^3+(n+1)^3=\frac{n^2(n+1)^2}{4}+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}=\frac{(n+1)^2((n+1)+1)^2}{4}$$
Hence, the given proposition is true by the principle of mathematical induction.
