Originally Posted by

**Archie Meade** Of course, that is not the complete proof.

The formula has been derived by examining a pattern from the first few terms of n.

To be sure whether this formula holds for all n natural numbers,

we try to establish that the formula being true for some n=k, __causes__ the formula to be true for the next n=k+1.

Because we use n or k, we are attempting to establish this term-by-term relationship in general __for all pairs of terms__.

If it works then a chain-reaction is established for all n.

To do this we try to prove

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

Therefore, we start by adding the next cube

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

$\displaystyle =\frac{(k+1)^2\left(4(k+1)+k^2\right)}{4}=\frac{(k +1)^2(4k+4+k^2)}{4}=\frac{(k+1)^2(k+2)^2}{4}$

Therefore the formula holds for all n, as it holds for n=1.