<s>[INST] Surprising Summation (1) $\sum_{i=1}^n(n-i+1)(2i-1)=\sum_{i=1}^n i^2$ Show that 

$$\sum_{i=1}^n(n-i+1)(2i-1)=\sum_{i=1}^n i^2$$

without expanding the summation to its closed-form solution, i.e. $\dfrac 16n(n+1)(2n+1)$ or equivalent.
E.g., if $n=5$, then
$$5(1) + 4(3)+ 3(5)+2(7)+1(9)=1^2+2^2+3^2+4^2+5^2$$
Background as requested:
The summands for both equations are not the same but the results are the same. The challenge here is to transform LHS into RHS without solving the summation. It seems like an interesting challenge.
Further edit
Thanks to those who voted to reopen the question!
 [/INST] $\begin{array}\\
\sum_{r=1}^n(n-r+1)(2r-1)
&=\sum_{r=1}^n(n+1)(2r-1)-\sum_{r=1}^nr(2r-1)\\
&=(n+1)n^2-2\sum_{r=1}^nr^2+\sum_{r=1}^nr\\
&=(n+1)n^2-2\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\\
&=\frac{6(n+1)n^2-2n(n+1)(2n+1)+3n(n+1)}{6}\\
&=n(n+1)\frac{6n-2(2n+1)+3}{6}\\
&=n(n+1)\frac{2n+1}{6}\\
&=\frac{n(n+1)(2n+1)}{6}\\
\end{array}
$
Now that's quite a surprise!
I'll stop here
a try to find a magic trick
later.
</s>