<s>[INST] Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is: 
$1 \times 1 + 2 \times 2 + 3 
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to 
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \times 4 + 11 \times 3 + 13 \times 2 + 15 \times 1 = 204$.
The second one logic is that I start with $1$, then I increment by $2$ each time and subtract $1$ from the second one, until I reach $1$. 
For $n = 2$.
$1 \times 1 + 2 \times 2 = 4 = 1 \times 2 + 3 \times 1$.
For $n = 3$.
$1 \times 1 + 2 \times 2 + 3 \times 3 = 1 \times 3 + 3 \times 2 + 5 \times 1$ 
The question is, why is it supposed to be the equal the one above? I tried it with a lot of values for $n$?
 [/INST] Alternatively:
$$\sum_{k=1}^n k^2=1^2+2^2+3^2+\cdots +n^2=\\
\color{red}1+(\color{red}1+\color{blue}3)+(\color{red}1+\color{blue}3+\color{green}5)+\cdots +(\color{red}1+\color{blue}3+\color{green}5+\cdots +\color{brown}{(2n-1)})=\\
\color{red}1\cdot n+\color{blue}3\cdot (n-1)+\color{green}5\cdot (n-2)+\cdots +\color{brown}{(2n-1)}\cdot 1.$$
Note: This is an algebraic presentation of Robert Z's wonderful PWW!
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