Using Riemann sum in a Combinatorial Proof Question The combinatorial proof is to prove that 
$$\sum_{k=1}^n k\cdot{n \choose k} = n\cdot 2^{n-1}$$
The solution says to count the leader and committee in two ways. 
The first way you choose the leader in n diff ways. Then choose the rest of the committee in $2^{n-1}$ ways to get the second side of the equation.
The second way you choose a committee with $k$ people in ${ n \choose k}$ ways, and then there are $k$ ways to choose its leader. 
So this is suppose to give you the left side of the equation 
$$ \sum_{k=1}^n k\cdot{n \choose k} = n\cdot 2^{n-1}$$
What I'm not understanding is why in the second part, you use a Riemann's sum to add all the ways. Because if you're just choosing $k$ people, I'm not sure why the $k$ value has to increase.
Thanks.
 A: As was noted in the comments, this has nothing to do with Riemann sums; I will ignore the term and assume that you’re really asking why we’re summing over $k=1,2,\ldots,n$. I’ll try to answer this by expanding on the explanation that you already have, adding more detail.
You have a pool of $n$ people; from that pool you’re to form a committee and choose one of its members to lead it. The committee may be of any size from $1$ through $n$. (It has to have at least one member, because you have to choose one member to be its leader.) The problem is to count the number of different ways in which this can be done.
We could first choose the leader; there are $n$ ways to do that, since any member of the pool can be the leader. Then we can choose any subset (including the empty subset) of the remaining $n-1$ people to fill out the rest of the committee. A set with $n-1$ members has $2^{n-1}$ different subsets, so there are $2^{n-1}$ ways to choose the rest of the committee. Thus, there are altogether $n2^{n-1}$ ways to choose a non-empty committee and designate one of its members the leader.
However, we can also count the committees (with designated leader) of each possible size and add these numbers to get the total number of possibilities. Suppose that $1\le k\le n$; how many committees (with designated leader) are there of size $k$? The pool of $n$ available individuals has $\binom{n}k$ subsets of size $k$, so there are $\binom{n}k$ possible committees of size $k$. Once we’ve chosen one of these committees, we can choose any one of its $k$ members to be its leader, so there are altogether $k\binom{n}k$ ways to choose a committee of size $k$ and designate its leader. But we want the number of ways to choose a committee of any possible size and designate its leader. For each $k$ from $1$ through $n$ there are $k\binom{n}k$ ways to choose a committee of size $k$ and pick its leader, so to get the grand total we must add these numbers, getting
$$\sum_{k=1}^nk\binom{n}k\;.$$
Since this grand total is just the number of ways to form a committee of any available size and pick its leader, which we already saw is $n2^{n-1}$, we now know that
$$\sum_{k=1}^nk\binom{n}k=n2^{n-1}\;.$$
A: The idea of a Riemann Sum does not seem to apply to this question. However, the sum on the left side is computing the number of ways to choose a team of $k$ people and a leader from within that team. The team can have anywhere from $1$ to $n$ members, thus we need to sum for all values of $k$ from $1$ to $n$ to capture all the possibilities for the size of the team.
