# help use induction to prove

• Nov 29th 2012, 03:14 AM
harryisland
help use induction to prove
use induction to prove

13 + 23 + 33 + . . . + n3 = (1 + 2 + 3 + . . . + n)2
• Nov 29th 2012, 03:43 AM
Re: help use induction to prove
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• Nov 29th 2012, 05:35 AM
topsquark
Re: help use induction to prove
Quote:

Originally Posted by harryisland
use induction to prove

13 + 23 + 33 + . . . + n3 = (1 + 2 + 3 + . . . + n)2

I am going to assume that we are aware of the identity
$\sum_{i = 1}^k = \frac{k(k + 1)}{2}$

It is not necessary, but useful.

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

So assume that there exists a n = k such that
$1^3 + 2^3 + ~...~ + k^3 = (1 + 2 + ~...~ + k)^2$

So what can we do with n = k + 1?
$(1^3 + 2^3 + ~...~ + k^3) + (k + 1)^3 = ((1 + 2 + ~...~ + k) + (k + 1))^2$

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

The first terms on both sides cancel, leaving
$(k + 1)^3 = 2(1 + 2 + ~...~ + k)(k + 1) + (k + 1)^2$

I'm going to leave the rest to you, but with the reminder of the sum above:
$\sum_{i = 1}^k = 1 + 2 + ~...~ + k = \frac{k(k + 1)}{2}$

If you need clarification, please let us know.

-Dan