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 $n$?

The first sum is $\sum _{n=1} ^k n^2 $ , while the second sum is $\sum _{n=1} ^k (2n-1)(k+1-n)$ . 
Let us try to relate the two. We have :-
$$\sum _{n=1} ^k(2n-1)(k+1-n) =k\sum _{n=1} ^k(2n-1) -2\cdot\sum_{n=1} ^kn^2+\sum_{n=1}^kn+\sum_{n=1}^k2n-1$$
$$=2k\cdot\frac{k\cdot(k+1)}{2} - k^2 -2\sum_{n=1}^kn^2+\frac{k\cdot(k+1)}{2} +2\cdot\frac{k(k+1)}{2}-k= \frac{k\cdot(k+1)\cdot(2k+1)}{2}-2\sum _{n=0}^kn^2= 3\cdot \sum _{n=0}^kn^2-2\sum _{n=0}^kn^2=\boxed {\sum _ {n=1} ^k n^2} $$
