Mathematical Induction help

**tokio** · March 5th 2009, 08:41 AM

1^3+2^3+...+n^3 = [n(n+1)/2]^2 n>= 1

If anyone has a link to write in mathematical nottaion on this forum, i will appreciate it, thanks.

**running-gag** · March 5th 2009, 08:50 AM

Hi

Have a look to the LATEX tutorial in the LATEX forum

Let P(n) be $1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$

Check that P(1) is true

Suppose that P(k) is true and show that P(k+1) is true

**tokio** · March 5th 2009, 09:25 AM

Thanks for the help, but i need help on the induction stage, when i do n+1. Here is what i have.

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

**running-gag** · March 5th 2009, 10:38 AM

Originally Posted by tokio

Thanks for the help, but i need help on the induction stage, when i do n+1. Here is what i have.

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

You suppose that P(n) is true, which means that $1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$

You want to prove that P(n+1) is true : $1^3+2^3+...+n^3+(n+1)^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$

$1^3+2^3+...+n^3+(n+1)^3 =(1^3+2^3+...+n^3)+(n+1)^3 =\left(\frac{n(n+1)}{2}\right)^2+ (n+1)^3$

Factor (n+1)²/4

**tokio** · March 6th 2009, 09:58 AM

Originally Posted by running-gag

You suppose that P(n) is true, which means that $1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$

You want to prove that P(n+1) is true : $1^3+2^3+...+n^3+(n+1)^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$

$1^3+2^3+...+n^3+(n+1)^3 =(1^3+2^3+...+n^3)+(n+1)^3 =\left(\frac{n(n+1)}{2}\right)^2+ (n+1)^3$

Factor (n+1)²/4

Im still stuck on the induction stage, It seems like if i still factor that i still cant get both sides of the quation to be logically equilvalent.

**Grandad** · March 6th 2009, 11:21 AM

Hello tokio

Originally Posted by tokio

Im still stuck on the induction stage, It seems like if i still factor that i still cant get both sides of the quation to be logically equilvalent.

Here's the bit you need:

$\left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3$

$= \frac{n^2\color{red}(n+1)^2}{\color{red}4} + \frac{4(n+1)\color{red}(n+1)^2}{\color{red}4}$

$= \color{red}\frac{(n+1)^2}{4}\color{black}(n^2 + 4(n+1))$

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

Can you complete it now?

Grandad

**tokio** · March 6th 2009, 06:22 PM

Originally Posted by Grandad

Hello tokioHere's the bit you need:

$\left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3$

$= \frac{n^2\color{red}(n+1)^2}{\color{red}4} + \frac{4(n+1)\color{red}(n+1)^2}{\color{red}4}$

$= \color{red}\frac{(n+1)^2}{4}\color{black}(n^2 + 4(n+1))$

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

Can you complete it now?

Grandad

Thanks for the help but i still cant believe im confused. Im assume that your just working onthe right hand side, i dont see where you proved for (n+1).

**Grandad** · March 6th 2009, 10:08 PM

Hello tokio

Originally Posted by tokio

Thanks for the help but i still cant believe im confused. Im assume that your just working onthe right hand side, i dont see where you proved for (n+1).

That's correct. I thought it was just the manipulation of the RHS that you couldn't do. The complete proof is:

Let $P(n)$ be the propositional function: $1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$

Then $P(n) \Rightarrow 1^3+2^3+...+n^3+(n+1)^3 =(1^3+2^3+...+n^3)+ (n+1)^3 =\left(\frac{n(n+1)}{2}\right)^2+ (n+1)^3$

$= \dots$ (as in my previous posting)

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

$= \left(\frac{(n+1)([n+1]+1)}{2}\right)^2$

$\Rightarrow P(n+1)$

$P(1)$ is $1^3 = \left(\frac{1 \cdot 2}{2} \right)^2$ , which is true. Hence $P(n), \forall n \in \mathbb{N}$ .

Grandad

PS If you want to see the LaTeX that's been used to generate any of the expressions, just click on them and a little window will open showing you the code.

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