<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] 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 $n=k+1$:
$\sum_{k=1}^{n+1} (2k-1)((n+1)-k+1) \\
=\sum_{k=1}^{n+1} (2k-1)((n-k+1) + \sum_{k=1}^{n+1} (2k-1) \\
=\sum_{k=1}^{n} (2k-1)((n-k+1) + (2(n+1)-1)(n-(n+1)+1) + \sum_{k=1}^{n+1} (2k-1) \\
=\sum_{k=1}^{n} k^2 +(2n+1)\times0 + (n+1)^2 \\
=\sum_{k=1}^{n+1} k^2$
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