TITLE: Formal notation of the sum of the first $n$ squares
QUESTION [8 upvotes]: I've found the following summation to be true
$$\sum_{i=1}^3 i^2 = 3+(3+2)+(3+2+1)$$ 
and it works for the first $n$ numbers, I also know that there is already a summation for that: $\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$ but that's not what I'm looking to use, my question is how would I appropiately write that first summation, could it be the following? 
$$\sum_{i=1}^n i^2 = n+\big(n+(n-1)\big)+\big(n+(n−1)+(n−2)\big)+\dots$$
I think that looks kind of ugly. 
How would I write the previous summation with a correct notation?

REPLY [2 votes]: Putting aside all the neat manipulations one can do with summation identities, note that
$$\begin{align}
\text{RHS}=\quad
&\color{red}3\\
+(&\color{red}3+\color{green}2)+\\
+(&\color{red}3+\color{green}2+\color{blue}1)\\\\\
=3(&\color{red}3)+2(\color{green}2)+1(\color{blue}1)\\\\
=\; \; \; &3^2+2^2+1^2\\\\
=\;\;\;&\sum_{i=1}^3 i^2 = \text{LHS}
\end{align}$$
You can prove the general case using the same method.   

The formal notation for $n$ numbers, as requested, and the proof that it is equal to the sum of squares is as follows:
$$\begin{align}
\color{purple}{\sum_{i=1}^n}\sum_{j=1}^i \color{orange}{(n+1-j)}
&=\color{purple}{\sum_{i=1}^n}\sum_{j=1}^i\color{orange}{\sum_{k=j}^n1}\\
&=\sum_{j=1}^n\color{purple}{\sum_{i=j}^n}\color{orange}{\sum_{k=j}^n1}
&& (1\le j\le i\le n)\\
&=\sum_{r=1}^n \;\;\color{purple}{\sum_{i=(n+1-r)}^n} \;\color{orange}{\sum_{k=(n+1-r)}^n 1}
&&(r=n+1-j)\\
&=\sum_{r=1}^n \qquad \color{purple}r\;\;\cdot \quad \color{orange}r\\
&=\sum_{r=1}^n r^2 =\sum_{i=1}^n i^2
\end{align}$$