I was trying to show that $\sum \limits_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ but instead I got this $[\frac{n(n+1)}{2}]^2$ which from my understanding I basically proved another summation formula which is $\sum \limits_{k=1}^n k^3$. Obviously I must have done something wrong. So I am going to show you how I got the summation formula wrong.
$s_n = 1^2 + 2^2 + ... + (n-1)^2 + n^2 $
In reverse order
$s_n = n^2 + (n-1)^2 + ... + 2^2 + 1^2 $
I decided to square root the partial sum which probably what led to the wrong answer. But I do not know, I am clumsy when I write on paper.
$\sqrt{s_n} = 1 + 2 + ... + (n-1) + n$
In reverse order
$\sqrt{s_n} = n + (n-1) + ... + 2 + 1$
Adding the two partial sums
$\sqrt{s_n} + \sqrt{s_n} = 2\sqrt{s_n}$
$2\sqrt{s_n} = (n+1) + (n+1) + ...$
$2\sqrt{s_n} = n(n+1)$
$\sqrt{s_n} = \frac{n(n+1)}{2}$
$(\sqrt{s_n})^2 = [\frac{n(n+1)}{2}]^2$
Now my question what did I do wrong. Can somebody show me the correct way. I am pretty sure this a fake proof, or a minor error. Thank you.
P.S. I am no latex expert and this not homework just for practice.