Surprising Summation (4): $\frac 12 \sum_{i=1}^n (n+1-i)(n+i)=\sum_{i=1}^n i^2$ 
Show that
  $$\frac 12 \sum_{i=1}^n (n+1-i)(n+i)=\sum_{i=1}^n i^2$$
  without expanding the summation to its closed form, i.e. $\dfrac 16n(n+1)(2n+1)$ or equivalent.

e.g.for $n=5$,
$$\frac12\bigg[5(6)+4(7)+3(8)+2(9)+1(10)\bigg]=1^2+2^2+3^2+4^2+5^2 $$
(Note: Whilst looking through similar questions posted earlier, another form of the summation cropped up and I thought I would post it as a question.)
 A: The inductive proof is fairly short, and avoids the closed form: replacing $n$ with $n+1$ increases each summand on the left-hand side by $(n+1-r) + (n+r) + 1 = 2n+2$ (since $(a+1)(b+1) - ab = a+b+1$), and introduces a new final summand of $2n+2$.
The overall increase is $\frac{1}{2} (n(2n+2) + 2n+2) = n^2 + 2n + 1 = (n+1)^2$, equal to the new final summand on the right-hand side.
A: This answer should not be taken too seriously, since usually we would calculate the series much more efficiently. But since OPs is asking for alternatives ... :-)
Here we show the validity of the identity by deriving a generating function valid for LHS and RHS.

The following is valid
  \begin{align*}
\sum_{n=0}^{\infty}\left(\sum_{r=1}^nr^2\right)x^n
&=\sum_{n=0}^\infty\left(\frac{1}{2}\sum_{r=1}^{n}(n+1-r)(n+r)\right)x^n\\
&=\frac{x(1+x)}{(1-x)^4}
\end{align*}

We start with the simpler series:

We obtain
  \begin{align*}
\sum_{n=0}^{\infty}\sum_{r=1}^nr^2x^n&=\frac{1}{1-x}\sum_{n=0}^{\infty}n^2x^n\tag{1}\\
&=\frac{1}{1-x}\left(xD_x\right)^2\sum_{n=0}^\infty x^n\tag{2}\\
&=\frac{1}{1-x}\left(xD_x\right)^2\frac{1}{1-x}\tag{3}\\
&=\frac{x(1+x)}{(1-x)^4}
\end{align*}

Comment: 


*

*In (1) we use $\frac{1}{1-x}$ the relationship 
\begin{align*}
\frac{1}{1-x}\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty\sum_{k=0}^na_k x^n
\end{align*}

*In (2) we use  the differential Operator $D_x$ and the relationship
\begin{align*}
\sum_{n=0}^{\infty}na_nx^n=\left(xD_x\right)\sum_{n=0}^{\infty}a_nx^n
\end{align*}

*In (3) we use the geometric series expansion

*In (4) we apply the $(xD_x)$ operator successively. We use here in the following
\begin{align*}
\left(xD_x\right)\frac{1}{1-x}&=\frac{x}{(1-x)^2}\\
\left(xD_x\right)^2\frac{1}{1-x}&=\frac{x(1+x)}{(1-x)^3}\\
\left(xD_x\right)^3\frac{1}{1-x}&=\frac{x(1+4x+x^2)}{(1-x)^4}\\
\end{align*}
Now the generating series for the other series:

We obtain
  \begin{align*}
\sum_{n=0}^{\infty}&\frac{1}{2}\sum_{r=1}^n(n+1-r)(n+r)x^n\\
&=\frac{1}{2}\sum_{n=0}^{\infty}\sum_{r=1}^nn(n+1)x^n+\frac{1}{2}\sum_{n=0}^{\infty}\sum_{r=1}^nr(1-r)x^n\\
&=\frac{1}{2}\sum_{n=0}^{\infty}(n^3+n^2)x^n+\frac{1}{2}\cdot\frac{1}{1-x}\sum_{n=0}^{\infty}n(n-1)x^n\\
&=\frac{1}{2}\left(\left(xD_x\right)^3\frac{1}{1-x}+\left(xD_x\right)^2\frac{1}{1-x}\right)\\
&\qquad\qquad+\frac{1}{2}\left(\left(xD_x\right)\frac{1}{1-x}-\left(xD_x\right)^2\frac{1}{1-x}\right)\tag{5}\\
&=\frac{1}{2}\left(\frac{x(1+4x+x^2)}{(1-x)^4}+\frac{x(1+x)}{(1-x)^3}+\frac{x}{(1-x)^3}-\frac{x(1+x)}{(1-x)^4}\right)\\
&=\frac{1}{2}\left(\frac{3x^2+x^3}{(1-x)^4}+\frac{x^2+2x}{(1-x)^3}\right)\tag{6}\\
&=\frac{1}{2}\cdot\frac{2x^2+2x}{(1-x)^4}\\
&=\frac{x(1+x)}{(1-x)^4}
\end{align*}
and the claim follows.

Comment:


*

*In (5) we use the identities stated in the comment section above in (4).

*In (6) we collect according to the same denominator.
A: $$\begin{align}
\sum_{i=1}^n (n+1-i)(n+i)-\sum_{i=1}^ni^2
&=\sum_{i=1}^n (n+1-i)(n+i)-\sum_{i=1}^n(n+1-i)^2\\
&=\sum_{i=1}^n (n+1-i)(2i-1)\\
&=\sum_{i=1}^n\sum_{j=i}^n (2i-1)\\
&=\sum_{j=1}^n\sum_{i=1}^j (2i-1)
&&(1\le i\le j\le n)\\
&=\sum_{j=1}^n\sum_{i=1}^j i^2-(i-1)^2\\
&=\sum_{j=1}^n j^2\\
&=\sum_{i=1}^n i^2\\
\Rightarrow \frac 12\sum_{i=1}^n (n+1-i)(n+i)
&=\sum_{i=1}^n i^2\qquad\blacksquare
\end{align}$$
