Q:

Summation involving binomial coefficients #1

I am asked to find the following sum:$$\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}\frac{(k-i)}{M-100}$$
The first two terms in the expression i.e, the binomial coefficients is actually $\binom{M}{100}$ as $\binom{k}{i}$ is the coefficient of $x^i$ in $(1+x)^k$ and $\binom{M-k}{100-i}$ is the coefficient of $x^{100-i}$ in $(1+x)^{M-k}$. 
Simplifying things I get an answer which is no was close to the actual answer.

A:

Note that by Vandermonde's identity,
$$\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}k=k\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}=k\binom{M}{100}$$
and
$$\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}i=\sum_{i=1}^{100}\frac{k}{i}\binom{k-1}{i-1}\binom{M-k}{100-i}i=k\sum_{i=1}^{100}\binom{k-1}{i-1}\binom{M-k}{99-(i-1)}\\
=k\sum_{j=0}^{99}\binom{k-1}{j}\binom{M-k}{99-j}=k\binom{M-1}{99}=\frac{k100}{M}\binom{M}{100}.$$
Hence 
$$\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}\frac{(k-i)}{M-100}=
\frac{1}{M-100}\left(k\binom{M}{100}-\frac{k100}{M}\binom{M}{100}\right)=
\frac{k}{M}\binom{M}{100}.$$

