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.

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- Mar 5th 2009, 08:41 AMtokioMathematical 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, 08:50 AMrunning-gag
Hi

Have a look to the LATEX tutorial in the LATEX forum

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

Check that P(1) is true

Suppose that P(k) is true and show that P(k+1) is true - Mar 5th 2009, 09:25 AMtokio
Thanks for the help, but i need help on the induction stage, when i do n+1. Here is what i have.

$\displaystyle \frac{n(n+1)^2+2(n+1)^3}{2} = \frac{((n+1(n+2)))}{2}^2$ - Mar 5th 2009, 10:38 AMrunning-gag
You suppose that P(n) is true, which means that $\displaystyle 1^3+2^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$

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

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

Factor (n+1)²/4 - Mar 6th 2009, 09:58 AMtokio
- Mar 6th 2009, 11:21 AMGrandadProof by induction
Hello tokioHere's the bit you need:

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

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

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

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

Can you complete it now?

Grandad - Mar 6th 2009, 06:22 PMtokio
- Mar 6th 2009, 10:08 PMGrandadProof by induction
Hello tokioThat's correct. I thought it was just the manipulation of the RHS that you couldn't do. The complete proof is:

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

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

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

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

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

$\displaystyle \Rightarrow P(n+1)$

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

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.