How to reason that the following sum is zero? Assume the following summation,
$$
\sum_{i=0}^{1000}\left(-1\right)^{i}{1000 \choose i}\left(100 - i\right)^{500}.
$$
I know that this summation is zero, $0$ ( I've checked it with Maple, though ). But I cannot find any proof for that!. Can you provide any help ?.
P.S. This is not a homework problem, just a question one of my friend asked me.
 A: If $(Δ p)(x)=p(x+1)-p(x)$ denotes the step-one difference operator, your expression is equal to
$$0=(Δ^{1000} p)(x)\quad\text{where}\quad p(x)=(100-x)^{500}$$
Since each application of $Δ$ lowers the degree of the polynomial by one, already $(Δ^{501} p)(x)=0$, and moreso the higher order differences.

If $p(x)=x^k$, then 
$$
(Δ p)(x)=p(x+1)-p(x)=(x+1)^k-x^k=(x+1)^{k-1}+(x+1)^{k-2}x+...+(x+1)x^{k-2}+x^{k-1}
$$
is a polynomial of one degree less, this extends linearly to all polynomials. Thus $\deg(Δ^m p)=\deg p-m$ as long as $m\le\deg p$, and $Δ^{1+\deg p} p=0$.
For the higher order differences on gets the formula
$$
(Δ^m p)(x)=\sum_{i=0}^m(-1)^{m-i}\binom{m}{i} p(x+i)
$$
This is obviously true for $m=0$, and easy to check for $m=1$, let's do the induction step
\begin{align}
(Δ^{m+1} p)(x)&=(Δ^m p)(x+1)-(Δ^m p)(x)
\\
&=\sum_{i=0}^m(-1)^{m-i}\binom{m}{i} p(x+i+1)-\sum_{i=0}^m(-1)^{m-i}\binom{m}{i} p(x+i)
\\
&=\sum_{i=1}^{m+1}(-1)^{m-i+1}\binom{m}{i-1} p(x+i)+\sum_{i=0}^m(-1)^{m-i+1}\binom{m}{i} p(x+i)
\\
&=\sum_{i=0}^{m+1}(-1)^{m+1-i}\binom{m+1}{i} p(x+i)
\end{align}

Even easier is the derivation using the translation operator $(Tp)(x)=p(x+1)$. Then $Δ=(T-I)$ and 
$$
Δ^m=(T-I)^m=\sum_{i=0}^m\binom{m}{i}(-1)^{m-i}T^{i}.
$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{i=0}^{1000}\pars{-1}^{i}{1000 \choose i}\pars{100 - i}^{500} = 0:\
     {\large ?}}$

\begin{align}
&\sum_{i=0}^{1000}\pars{-1}^{i}{1000 \choose i}\pars{100 - i}^{500}
=\sum_{i=0}^{1000}\pars{-1}^{i}{1000 \choose i}\sum_{\ell = 0}^{500}
{500 \choose \ell}100^{500 - \ell}\pars{-i}^{\ell}
\\[3mm]&=\sum_{\ell = 0}^{500}{500 \choose \ell}\pars{-1}^{\ell}100^{500 - \ell}
\underbrace{\sum_{i=0}^{1000}{1000 \choose i}\pars{-1}^{i}i^{\ell}}
_{\ds{=\ 0;\quad\mbox{See bellow}}}
\end{align}

\begin{align}
\pars{1 - z}^{1000}&=\sum_{i = 0}^{1000}{1000 \choose i}\pars{-1}^{i}z^{i}
\\
-1000z\pars{1 - z}^{999}&=\sum_{i = 0}^{1000}{1000 \choose i}\pars{-1}^{i}iz^{i}
\\
-1000z\pars{1 - z}^{999} + 1000\times999 z^{2}\pars{1 - z}^{998}&
=\sum_{i = 0}^{1000}{1000 \choose i}\pars{-1}^{i}i^{2}z^{i}\,,
\end{align}
If we continue this procedure we get increasing powers of $i$ ( up to $i^{\ell}$ with $\ell \leq 500$ ) in the right member.
Then, set $z = 1$ in both members and we'see that all the sums vanish out.
