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 $n$?
 A: Visual proof of
$$\sum_{k=1}^{n}k^2=\sum_{k=1}^{n} (2k-1)(n+1-k).$$
Take a look at this picture  for $n=5$. 

$$1+2^2+3^2+4^2+5^2=\underbrace{1\cdot 5}_{k=1}+\underbrace{3\cdot 4}_{k=2} +\underbrace{5\cdot  3}_{k=3}+ \underbrace{7\cdot 2}_{k=4}+ \underbrace{9\cdot 1}_{k=5}.$$
A: 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 :-
$$\sum _{n=1} ^k(2n-1)(k+1-n) =k\sum _{n=1} ^k(2n-1) -2\cdot\sum_{n=1} ^kn^2+\sum_{n=1}^kn+\sum_{n=1}^k2n-1$$
$$=2k\cdot\frac{k\cdot(k+1)}{2} - k^2 -2\sum_{n=1}^kn^2+\frac{k\cdot(k+1)}{2} +2\cdot\frac{k(k+1)}{2}-k= \frac{k\cdot(k+1)\cdot(2k+1)}{2}-2\sum _{n=0}^kn^2= 3\cdot \sum _{n=0}^kn^2-2\sum _{n=0}^kn^2=\boxed {\sum _ {n=1} ^k n^2} $$
A: 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!
A: We have $$\sum_{k=1}^j (2k-1)\cdot (j+1-k)=(j+1)\sum_{k=1}^j (2k-1)-2\sum_{k=1}^j k^2+\sum_{k=1}^j k$$ $$=(j+1)j^2-\frac{j(j+1)(2j+1)}{3}+\frac{j(j+1)}{2}$$ $$=\frac{j(j+1)(2j+1)}{6}=\sum_{k=1}^j k^2 $$ showing that the equality is no coincidence , but holds for every positive integer $j$.
A: Proof by induction that $\sum_{k=1}^{n} (2k-1)(n-k+1) = \sum_{k=1}^{n} k^2 \space \forall n \ge 1$:
Base case: For $n=1$ we have $1 \times 1 = 1$
Assume true for $n=k$.
For $n=k+1$:
$\sum_{k=1}^{n+1} (2k-1)((n+1)-k+1) \\
=\sum_{k=1}^{n+1} (2k-1)((n-k+1) + \sum_{k=1}^{n+1} (2k-1) \\
=\sum_{k=1}^{n} (2k-1)((n-k+1) + (2(n+1)-1)(n-(n+1)+1) + \sum_{k=1}^{n+1} (2k-1) \\
=\sum_{k=1}^{n} k^2 +(2n+1)\times0 + (n+1)^2 \\
=\sum_{k=1}^{n+1} k^2$
A: Nice observation!
It is indeed true that $ \sum_{k=0}^{n - 1} (2 k + 1) (n - k) = \sum_{k=1}^{n} k^2$.
This can be proved by using the formulas for $\sum 1, \sum k, \sum k^2$.
A: In you increase $n$ by $1$, all terms increase by $2k+1$ and there comes an extra term $2n+1$. Then $k^2+2k+1=(k+1)^2$.
