Sum first $n$ squares I am trying to find the formula for the sum of square number but I am struggle 
with it. I now that it isn't complex but I am young in maths.
What I've done:
Using odd numbers to find the pattern 
Using triangle number and the relative
 formula:
       $$\frac{n(n+1)}{2}$$
1                  $\rightarrow$
1+3               $\rightarrow$
1+3+5               $\rightarrow$
.
.
.
1+3+5+...=$n^2$
1+1+1+...=$n-1$
Thanks

What you can do is follow Euler's method : 
\begin{equation*}
\sum_{k=1}^n ((k+1)^3 - k^3) = (n+1)^3  - 1
\end{equation*}
But also : 
\begin{equation*}
\sum_{k=1}^n ((k+1)^3 - k^3) = \sum_{k=1}^n (3k^2 + 3k +1) 
\end{equation*}
Thus, knowing $\sum_{k=1}^n k = \frac{n(n+1)}{2}$, we get : 
\begin{equation*}
3\sum_{k=1}^n k^2 = (n+1)^3 - 1 - 3 \frac{n(n+1)}{2} - n = \frac{(2n+1)n(n+1)}{2}
\end{equation*}
thus: 
\begin{equation*}
\sum_{k=1}^n k^2 = \frac{(2n+1)n(n+1)}{6}
\end{equation*}
