proof of expected value of the hypergeometric distribution

We will first prove a useful property of binomial coefficients. We know

{n\\choose k}=\\frac{n!}{k!(n-k)!}.

This can be transformed to

{n\\choose k}=\\frac{n}{k}\\frac{(n-1)!}{(k-1)!(n-1-(k-1))!}=\\frac{n}{k}{n-1%\\choose k-1}.

(1)

Now we can start with the definition of the expected value:

E[X]=\\sum_{x=0}^{n}\\frac{x{K\\choose x}{M-K\\choose n-x}}{{M\\choose n}}.

Since for $x=0$ we add a $0$ in this we can say

E[X]=\\sum_{x=1}^{n}\\frac{x{K\\choose x}{M-K\\choose n-x}}{{M\\choose n}}.

Applying equation (1) we get:

E[X]=\\frac{nK}{M}\\sum_{x=1}^{n}\\frac{{K-1\\choose x-1}{M-1-(K-1)\\choose n-1-(x-%1)}}{{M-1\\choose n-1}}.

Setting $l:=x-1$ we get:

E[X]=\\frac{nK}{M}\\sum_{l=0}^{n-1}\\frac{{K-1\\choose l}{M-1-(K-1)\\choose n-1-l}}%{{M-1\\choose n-1}}.

The sum in this equation is $1$ as it is the sum over all probabilities of a hypergeometric distribution. Therefore we have

E[X]=\\frac{nK}{M}.