TITLE: Why does $\sum\limits_{i=0}^{100} \binom{100}{i} \sum\limits_{j=i+1}^{101}\binom{101}{j}$ equal $2^{200}$?
QUESTION [6 upvotes]: According to WolframAlpha, the following is true:
$$\sum_{i=0}^{100} \frac{ \binom{100}{i} }{2^{100}}\sum_{j=i+1}^{101}\frac{ \binom{101}{j} }{2^{101}} = \frac{1}{2}$$
Can anyone tell me why that is?
I'm quessing it has something to do with the hypergeometric function.

REPLY [3 votes]: Note that 
$\displaystyle \binom ni\binom {n+1}j=\binom n{n-i}\binom {n+1}{n+1-j}$.
Hence
$$\begin{align}
&\sum_{i=0}^n\binom ni\sum_{j=i+1}^{n+1}\binom {n+1}j\\
&=\sum_{i=0}^n\sum_{j=i+1}^{n+1}\binom ni\binom {n+1}j\\
&=\frac 12\left[\sum_{i=0}^n\sum_{j=i+1}^{n+1}\binom ni\binom {n+1}j+\binom n{n-i}\binom {n+1}{n+1-j}\right]\\
&=\frac 12\left[\sum_{i=0}^n\sum_{j=i+1}^{n+1}\binom ni\binom {n+1}j+\sum_{r=0}^n\sum_{s=0}^r \binom nr\binom {n+1}s\right]
&&\scriptsize r=n-i;\atop \scriptsize s=n+1-j\\
&=\frac 12\left[\sum_{i=0}^n\sum_{j=i+1}^{n+1}\binom ni\binom {n+1}j+\sum_{i=0}^n\sum_{j=0}^i \binom ni\binom {n+1}j\right]\\
&=\frac 12\sum_{i=0}^n\sum_{j=0}^{n+1}\binom ni\binom {n+1}j\\
&=\frac 12\sum_{i=0}^n \binom ni\sum_{j=0}^{n+1}\binom {n+1}j\\
&=\frac 12\cdot 2^n\cdot 2^{n+1}\\
&=\color{red}{2^{2n}}
\end{align}$$
Putting $n=100$ gives the proof required.