use induction to prove

1^{3}+ 2^{3}+ 3^{3}+ . . . + n^{3}= (1 + 2 + 3 + . . . + n)^{2 }

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- Nov 29th 2012, 03:14 AMharryislandhelp use induction to prove
use induction to prove

1^{3}+ 2^{3}+ 3^{3}+ . . . + n^{3}= (1 + 2 + 3 + . . . + n)^{2 } - Nov 29th 2012, 03:43 AMLinkRe: help use induction to prove
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- Nov 29th 2012, 05:35 AMtopsquarkRe: help use induction to prove
I am going to assume that we are aware of the identity

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

It is not necessary, but useful.

First, does this hold for n = 1? Obviously.

So assume that there exists a n = k such that

$\displaystyle 1^3 + 2^3 + ~...~ + k^3 = (1 + 2 + ~...~ + k)^2$

So what can we do with n = k + 1?

$\displaystyle (1^3 + 2^3 + ~...~ + k^3) + (k + 1)^3 = ((1 + 2 + ~...~ + k) + (k + 1))^2$

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:

$\displaystyle (1 + 2 + ~...~ + k)^2 + (k + 1)^3 = (1 + 2 + ~...~ + k)^2 + 2(1 + 2 + ~...~ + k)(k + 1) + (k + 1)^2$

The first terms on both sides cancel, leaving

$\displaystyle (k + 1)^3 = 2(1 + 2 + ~...~ + k)(k + 1) + (k + 1)^2$

I'm going to leave the rest to you, but with the reminder of the sum above:

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

If you need clarification, please let us know.

-Dan