use induction to prove
13 + 23 + 33 + . . . + n3 = (1 + 2 + 3 + . . . + n)2
It is not necessary, but useful.
First, does this hold for n = 1? Obviously.
So assume that there exists a n = k such that
So what can we do with n = k + 1?
By our assumption the first k terms on the LHS can be replaced by ( )^2. We are also going to expand the RHS as a square:
The first terms on both sides cancel, leaving
I'm going to leave the rest to you, but with the reminder of the sum above:
If you need clarification, please let us know.