induction problem,

**craigmain** · June 7th 2009, 05:14 AM

I have been proving a couple of induction problems.
I am struggling when there are sequences on both sides of the equals.

How do I prove by induction.
$1^3+2^3+3^3+ ... +n^3={(1+2+3+...+n)}^2$

demonstrating for k=1 is trivial.
Assume for k, and then prove for (k+1).

$1^3+2^3+3^3+ ... +k^3+{(k+1)}^3={(1+2+3+...+k+(k+1))}^2$

using $\sum\limits_1^n n = \frac{n(n+1)}{2}$

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

I have tried expanding this, it quickly get's horrible, but also doesn't seem closer to solving the problem.

Thanks
Regards
Craig.

**Deadstar** · June 7th 2009, 05:57 AM

Since $1 + 2 + ... + n = \frac{1}{2}n(n+1)$
$(1 + 2 + ... + n)^2 = \bigg{(}\frac{1}{2}n(n+1)\bigg{)}^2$

So... You proved it for 1. Assume for n, prove for n+1

$1^3 + 2^3 + ... n^3 + (n+1)^3 = (1^3 + 2^3 + ... + n^3) + (n+1)^3$

$= (1 + 2 + ... + n)^2 + (n+1)^3$

$=\bigg{(}\frac{1}{2}n(n+1)\bigg{)}^2 + (n+1)^3$
$= \frac{1}{4}(n+1)^2(n^2 + 4n + 4)$
$= \bigg{(}\frac{1}{2}(n+1)(n+2)\bigg{)}^2$
$= (1+2+...+(n+1))^2$ .

Hence proved by induction

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