I have no idea how to go about it really. Any help is very much appreciated.

$
\sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}
$

2. Hi nunos

I suggest using mathematical induction

3. Originally Posted by songoku
Hi nunos

I suggest using mathematical induction
I already know about that, but thanks for the suggestion anyway. I am going to write here, what I have so far, and hopefully you or someone will point me in the right direction.

Proving for n = 1, easy.

Assuming it's true for n, let's evaluate for n+1.

$\sum_{k=1}^{n+1} k^{n+1} = \sum_{k=1}^n k^n + k^{n+1} = \frac{n(n+1)(2n+1)}{6} + k^{n+1}$

$\sum_{k=1}^{n+1} k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$

$\Rightarrow \frac{n(n+1)(2n+1)}{6} + k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$

$\Leftrightarrow n(n+1)(2n+1) + 6k^{n+1} = (n+1)(n+2)(2n+3)$

And I don't know how to go from here... I don't even know if what I've done is relevant...

4. Originally Posted by nunos
I already know about that, but thanks for the suggestion anyway. I am going to write here, what I have so far, and hopefully you or someone will point me in the right direction.

Proving for n = 1, easy.

Assuming it's true for n, let's evaluate for n+1.

$\sum_{k=1}^{n+1} k^{n+1} = \sum_{k=1}^n k^n + k^{n+1} = \frac{n(n+1)(2n+1)}{6} + k^{n+1}$

$\sum_{k=1}^{n+1} k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$

$\Rightarrow \frac{n(n+1)(2n+1)}{6} + k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$

$\Leftrightarrow n(n+1)(2n+1) + 6k^{n+1} = (n+1)(n+2)(2n+3)$

And I don't know how to go from here... I don't even know if what I've done is relevant...
I believe you have some errors here. I think what you want is

$\sum_{k=1}^{n+1} k^2 =\sum_{k=1}^{n} k^2 +(n+1)^2 = \frac{n(n+1)(2n+1)}{6} + (n+1)^2 = \frac{(n+1)(n+2)(2n+3)}{6}$

5. Originally Posted by nunos
I already know about that, but thanks for the suggestion anyway. I am going to write here, what I have so far, and hopefully you or someone will point me in the right direction.

Proving for n = 1, easy.

Assuming it's true for n, let's evaluate for n+1.

$\sum_{k=1}^{n+1} k^{n+1} = \sum_{k=1}^n k^n + k^{n+1} = \frac{n(n+1)(2n+1)}{6} + k^{n+1}$
No, not $k^{n+1}$! Your sum was $\sum_{k=1}^n k^2$. You want $\sum_{k=1}^{n+1} k^2= \sum_{k=1}^n k^2+ (k+1)^2$.

$\sum_{k=1}^{n+1} k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$

$\Rightarrow \frac{n(n+1)(2n+1)}{6} + k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}$

$\Leftrightarrow n(n+1)(2n+1) + 6k^{n+1} = (n+1)(n+2)(2n+3)$

And I don't know how to go from here... I don't even know if what I've done is relevant...

6. Originally Posted by HallsofIvy
No, not $k^{n+1}$! Your sum was $\sum_{k=1}^n k^2$. You want $\sum_{k=1}^{n+1} k^2= \sum_{k=1}^n k^2+ (k+1)^2$.
Just a small correction here (just want to make sure the OP isn't confused).

Should be:

$\sum_{k=1}^{n+1} k^2= \sum_{k=1}^n k^2+ (\textcolor{red}{n}+1)^2$

7. You're all correct. That was a mistake. I had the impression that something had to be wrong. I had now completed the proof.

Thanks.