Compute this sum: $\sum_{k=0}^{n} k \binom{n}{k}$? Compute this sum:
$$\sum_{k=0}^{n} k \binom{n}{k}.$$
I tried  but I got stuck.
 A: $$\large S=\sum_{k=0}^{n} k \binom{n}{k}$$
$$\large S=0\binom{n}{0}+1\binom{n}{1}+2\binom{n}{2}+..+(n-1)\binom{n}{n-1}+n\binom{n}{n}$$
$$\large S=n\binom{n}{n}+(n-1)\binom{n}{n-1}+(n-2)\binom{n}{n-2}+..+1\binom{n}{1}+0\binom{n}{0}$$
Adding the above equations,
$$\large 2S=n(\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+..+\binom{n}{n-1}+\binom{n}{n})$$
$$\large 2S=n2^n$$
$$\large S=n2^{n-1}$$
A: Here's alternative way to get the result. The first thing to note is that
$$\sum_{k=0}^nk\binom{n}{k}=\sum_{k=0}^nk\cdot\frac{n!}{(n-k)!k!}=\sum_{k=1}^nk\cdot\frac{n!}{(n-k)!k!},$$
because the term with $k=0$ is equal to $0$. Next, cancelling the factor $k$ we find that
$$\sum_{k=1}^nk\cdot\frac{n!}{(n-k)!k!}=\sum_{k=1}^n\frac{n!}{(n-k)!(k-1)!}=\sum_{k=1}^nn\cdot\frac{(n-1)!}{(n-k)!(k-1)!}.$$
This can be further simplified by taking the factor $n$ out, and setting $j:=k-1$ to get
$$\sum_{k=1}^nn\cdot\frac{(n-1)!}{(n-k)!(k-1)!}=n\cdot\sum_{k=1}^n\frac{(n-1)!}{(n-k)!(k-1)!}=n\cdot\sum_{j=0}^{n-1}\frac{(n-1)!}{(n-1-j)!j!}.$$
We can now finish by noting that the terms of this last sum are again binomial coefficients:
$$n\cdot\sum_{j=0}^{n-1}\frac{(n-1)!}{(n-1-j)!j!}=n\cdot\sum_{j=0}^{n-1}\binom{n-1}{j}=n\cdot2^{n-1}.$$
A: A more intuitive way of thinking about this is to ask, "Given n people, how many possible 'teams' of people are there, given that each team has a leader?".
So on one hand, if a team has $k$ people in it, then there are ${n}\choose{k}$ ways to pick those $k$ people, and any of those $k$ people can be leader, so there are $k  {{n}\choose{k}}$ possibilities for a team with k people with a leader. Summing up over $k$, this means there are $$\sum_{k=0}^{n} k {{n}\choose{k}}$$ ways of picking a team with a leader from $n$ people.
On the other hand, there are $n$ people. Pick one of them to be a leader ($n$ possibilities) and then of the other $n-1$ people, they're either in the team or they're not, so that gives us $2^{n-1}$ ways of picking them. Multiplying, this gives us $$n 2^{n-1}$$.
As these expressions represent the same quantity, they are equal.
A: Hint. Let $X\sim \text{Bin}(n,p)$. Then it is known that 
$$\Bbb EX = np$$
But of course, $\Bbb EX = \sum_{k=0}^n k \Bbb P(X=k) =  \sum_{k=0}^n k \binom{n}{k}p^k(1-p)^k$. So
$$np = \sum_{k=0}^n k \binom{n}{k}p^k(1-p)^k$$
Try finishing from here.
A: $$\sum_{k=0}^nk\binom{n}{k}=\sum_{k=1}^nk\binom{n}{k}=\sum_{k=1}^nn\binom{n-1}{k-1}=n\sum_{k=0}^{n-1}\binom{n-1}{k}=n2^{n-1}$$
Identities used:
1)$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}=\dfrac nk\dfrac{(n-1)!}{(k-1)!(n-k)!}=\dfrac nk\binom{n-1}{k-1}$$
2)$$\sum_{k=0}^n\binom nk=2^n$$
which can be proved by expanding $(1+1)^n$ with binomial theorem.
