Expectation of sum of elements from a subset Let's say we have the set $S = {\{1,2,...,n\}}$. We will create a subset from $S$ called $S_k$ containing $k$ elements chosen from set $S$ where $0 \le k \le n$.
Each of the $k$ elements can only be picked once and each pick of an element from $S$ is random following a uniform distribution from all remaining elements in $S$.
What's the expected value of the sum of $S_k$?
I know that each pick of an element is without replacement. Therefore, the first pick follows that each element has a probability $\frac{1}{n}$ of being chosen, the second pick has a probability of $\frac{1}{n-1}$, etc. Where would I go from here?
 A: The expected value for any of the $k$ selected numbers is 
$$\frac{1+\cdots + n}{n}=\frac{n+1}{2}$$
hence the expected value of the sum of $S_k$ is
$$k\left(\frac{n+1}{2}\right)$$
Note that "without replacement" or "with replacement" doesn't matter for this question. 

To help see it . . .

First suppose we sample $k$ elements with replacement. 

Then each selection has expected value
$$\frac{1+\cdots + n}{n}=\frac{n+1}{2}$$
hence, since expectations add, the expected value for the sum of the $k$-element sample is
$$k\left(\frac{n+1}{2}\right)$$
Next, suppose we sample $k$ elements without replacement.

Then the first element removed has expected value 
$$\frac{1+\cdots + n}{n}=\frac{n+1}{2}$$

hence, the expected sum of the remaining numbers is
$$(1+\cdots + n)-\frac{n+1}{2}=(n-1)\left(\frac{n+1}{2}\right)$$
so the expected value for a random selection of one remaining $n-1$ numbers is
$$\frac{(n-1)\left(\frac{n+1}{2}\right)}{n-1}=\frac{n+1}{2}$$
Thus, the second selected number has the same expected value as the first.

By extension of the same reasoning, each of the $k$ selections has expected value
$$\frac{n+1}{2}$$
hence, since expectations add, the expected value for the sum of the $k$-element sample is
$$k\left(\frac{n+1}{2}\right)$$
