# Sum of the first n cubes

1. Nov 7, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
Can someone help me derive the formula for the sum of the first n cubes from the formula for the sum of the first n integers that elucidates the reason why the former is the square of the latter?

2. Relevant equations

3. The attempt at a solution

2. Nov 7, 2007

### rock.freak667

Well you could consider the sum from n=1 to N of (n+1)^4-n^4 and then expand using the method of differences

3. Nov 7, 2007

### ehrenfest

You mean use the fact that it telescopes?

Then you just get N^4-1? How does that help?

4. Nov 8, 2007

### Gib Z

$$(n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1$$ by the Binomial theorem.

Sum BOTH sides, and the LHS should get (N+1)^4 actually. Do you see how the RHS will be expressions in terms of the first N integers to the power of 3, 2, 1 and 0? You already know the cases for 0, 1 and 2, now you can work out 3.