Can someone please explain to me how I did this summation formula wrong? 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.

$\sqrt{a^2+b^2}\ne a+b$ in general unless at least one of $a,b$ is $0$
If $s_n=1^2+2^n+\cdots+(n-1)^2+n^2,$
how can you write $s_n=1+2+\cdots+(n-1)+n?$
(1)One way to proof is :
$ (r+1)^3-r^3=3r^2+3r+1$
Put $r=0,1,2,\cdots,n-1,n$ and add to get 
$(n+1)^3=3S_n+3(1+2+3+\cdots+n)+n=3S_n+3\frac{n(n+1)}2+n$
So, $S_n=...$
(2)We can use induction too:
Let $S(m)= \frac{m(m+1)(2m+1)}6$
$S(1)=\frac{1\cdot\cdot3}6=1$ which is true.
So, $S(m+1)=S(m)+m+1$
$=\frac{m(m+1)(2m+1)}6+(m+1)^2=\frac{(m+1)\{(m+1)+1\}\{2(m+1)+1\}}6$  
So the proposition holds true for $n=m+1$ if it is true for $n=m.$
