Hello db5vry Originally Posted by

**db5vry** I want to prove $\displaystyle \sum r^3 = \frac{1}{4}n^2(n + 1)^2$ by induction for all positive integers n but I'm having a bit of trouble after a while:

for n = 1: $\displaystyle 1^3$ = $\displaystyle 1$ so the statement's true for n = 1.

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

and for n = k = 1:

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

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

I'm not sure how to make progress from here. Is there anyone who can help out? Thanks

Just be a bit more careful with the notation, and it will help you see what to do next. You should include the limits of the summation, so your induction hypothesis should be:

Let us assume that, for $\displaystyle n=k$$\displaystyle \sum_{r=1}^nr^3 = \tfrac{1}{4}n^2(n + 1)^2$

Therefore:$\displaystyle \sum_{r=1}^kr^3 = \tfrac{1}{4}k^2(k + 1)^2$

Then:$\displaystyle \sum_{r=1}^{k+1}r^3 =\sum_{r=1}^kr^3+(k+1)^3$$\displaystyle = \tfrac{1}{4}k^2(k + 1)^2+(k+1)^3$

$\displaystyle = (\tfrac{1}{4}k^2+[k+1])(k + 1)^2$

$\displaystyle = \tfrac{1}{4}(k^2+4[k+1])(k + 1)^2$

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

So when $\displaystyle n = k+1,\; \sum_{r=1}^nr^3 = \tfrac{1}{4}n^2(n + 1)^2$ is also true.

Since you have already shown that this is true for $\displaystyle n = 1$, this is all you need to complete the proof.

Grandad