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.
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
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
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.