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
 A: I assume your question is to find the sum of the first n square numbers, 
$1+2^2+...+n^2$. There are several ways to do this: we could use the notion of anti-differences from "discrete calculus," or we could guess, the formula and use induction, or ... there are many ways to do it. One method, called perturbation, is simplest because it is essentially just an algebraic trick. You can find this method, and 7 others(applied directly to your problem), in chapter 2 of Concrete Math, from page 42(perturbation is  43). I would try to explain, but really, Knuth, Graham, and Patashnik do better than I ever could(Concrete Math is a classic, and contains a lot of more advanced stuff, but what I'm pointing you to is clearly written and very understandable).
Link to Concrete Math: https://notendur.hi.is/pgg/(ebook-pdf)%20-%20Mathematics%20-%20Concrete%20Mathematics.pdf 
Or you could buy the book on amazon, some of the formulas aren't legible in that PDF. I assure you, it is well worth it.
Possibility 2. Are you asking how to prove that the sum of the first n odds is equal to $n^2$? This you can get through manipulation of sums: 
notice that any odd can be written as $2k-1$ for sum integer $k$, and the nth odd will be exactly $2n-1$. So we are looking to compute 
$\sum_{k=1}^{n} 2k-1 =  2*\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1$.
 
Then we use the triangle formula and that $1+...+1$ n times is n to find that this is  
$2*\frac{n(n+1)}{2} - n = n(n+1)-n=n^2+n-n=n^2$, as desired.
A: 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*}
