Taking spheres from urn - expected value sum of number In one urn we have $n$-spheres with numbers from $1$ to $n$. We are taking one of them. Let $k$ be a number of this sphere. Then we return this sphere to urn and we are taking (without returning) $k$ spheres. Let $E(n)$ be expected value  sum of numbers on these spheres. Find $E(n)$.
I suppose that $k$ is random variable. Expected value of $k$ is of course $\frac{n+1}{2}$. But I don't have idea what I shoul do next. I will grateful for your help. Thanks in advance!

The first number drawn is equally likely to be any one of $1,2,3,\dots, n$. So each number has probability $\frac{1}{n}$ of being drawn.
Suppose that we have drawn the number $k$. Then at the second stage we draw $k$ balls, say in some order. Let $X_i$ be the number on the $i$-th ball drawn in the second stage. Then by the linearity of expectation, 
$$E(X_1+X_2+\cdots +X_k)=E(X_1)+E(X_2)+\cdots +E(X_k)=k\cdot\frac{n+1}{2}.$$
So the conditional expectation of the second stage sum, given that the ball drawn in the first stage is $k$, is equal to
$$\frac{(k)(n+1)}{2}.$$
Thus the expectation of the second stage sum is
$$\frac{1}{n}\cdot \frac{(1)(n+1)}{2}+\frac{1}{n}\cdot \frac{(2)(n+1)}{2}+\cdots + \frac{1}{n}\cdot \frac{(n)(n+1)}{2}.\tag{1}$$
Using the fact that $1+2+3+\cdots +n=\frac{(n)(n+1)}{2}$, we can greatly  simplify Expression (1). 
