1. 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.

2. 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

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

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

4. 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

5. 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.

6. Proof by induction

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?

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).

8. Proof by induction

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