TITLE: Consider the 1000-element subsets
QUESTION [6 upvotes]: Consider all 1000-element subsets of the set $A = \{ 1, 2, 3, ... , 2015 \}$. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

It will be:
$$\frac{a_1 + a_2 + ... + a_{k}}{k}$$
Obviously, one set is: $\{1, 2, 3, ..., 1000\}$ another could be: $\{1, 3, ... , 1000,1001\}$.
There are $2015 - 1 + 1 = 2015$ elements in set $A$. So: 
$\binom{2015}{1000}$ of 1000 element subsets are present. 
This problem is very tough for me to solve right now...
HINTS only please! Thank you!!

REPLY [2 votes]: A possible hint :
There is 

$ 2014 \choose 999 $ subsets with 1 as the smallest element
$ 2013 \choose 999 $ subsets with 2 as the smallest element
$\cdots$
$ 2015-k \choose 999 $ subsets with k as the smallest element

So you're looking at 
$$\frac{p}{q} = \frac{ \sum_{k=1}^{1016} k {2015-k \choose 999} }{ 2015 \choose 1000} $$
Can you calculate this sum? 
$$\sum_{k=1}^{1016} k {2015-k \choose 999} $$