# Mathematical Induction help

• Mar 5th 2009, 09:41 AM
tokio
Mathematical Induction help
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.
• Mar 5th 2009, 09:50 AM
running-gag
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
• Mar 5th 2009, 10:25 AM
tokio
Thanks for the help, but i need help on the induction stage, when i do n+1. Here is what i have.

$\frac{n(n+1)^2+2(n+1)^3}{2} = \frac{((n+1(n+2)))}{2}^2$
• Mar 5th 2009, 11:38 AM
running-gag
Quote:

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
• Mar 6th 2009, 10:58 AM
tokio
Quote:

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.
• Mar 6th 2009, 12:21 PM
Proof by induction
Hello tokio
Quote:

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?

• Mar 6th 2009, 07:22 PM
tokio
Quote:

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?

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).
• Mar 6th 2009, 11:08 PM
Proof by induction
Hello tokio
Quote:

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 $P(n), \forall n \in \mathbb{N}$.