Induction proof of summation

**db5vry** · January 21st 2010, 09:35 AM

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

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

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

and for n = k = 1:

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

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

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

**qmech** · January 21st 2010, 09:48 AM

You need to show:
$<br /> <br /> \frac{1}{4}(k)^2(k + 1)^2 + (k + 1)^3 = <br /> \frac{1}{4}(k+1)^2(k + 2)^2<br />$

So the LHS is your original sum and the next cubic term, the RHS is
the same form as your expression, but for k+1

**Grandad** · January 21st 2010, 09:56 AM

Hello db5vry

Originally Posted by db5vry

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

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

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

and for n = k = 1:

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

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

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 $n=k$

$\sum_{r=1}^nr^3 = \tfrac{1}{4}n^2(n + 1)^2$

Therefore:

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

Then:

$\sum_{r=1}^{k+1}r^3 =\sum_{r=1}^kr^3+(k+1)^3$

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

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

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

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

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

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

Grandad

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