Hint proving this $\sum_{k=0}^{n}\binom{2n}{k}k=n2^{2n-1}$ I need hint proving this $$\sum_{k=0}^{n}\binom{2n}{k}k=n2^{2n-1}$$
 A: Use the identity $k\dbinom{n}k=n\dbinom{n-1}{k-1}$:
$$\sum_{k=0}^n\binom{2n}kk=2n\sum_{k=0}^n\binom{2n-1}{k-1}=2n\sum_{k=0}^{n-1}\binom{2n-1}k\;.$$
Then note that 
$$\sum_{k=0}^{n-1}\binom{2n-1}k=\sum_{k=0}^{n-1}\binom{2n-1}{2n-1-k}=\sum_{k=n}^{2n-1}\binom{2n-1}k\;,$$
and
$$\sum_{k=0}^{n-1}\binom{2n-1}k+\sum_{k=n}^{2n-1}\binom{2n-1}k=\sum_{k=0}^{2n-1}\binom{2n-1}k\;,$$
so
$$\sum_{k=0}^{n-1}\binom{2n-1}k=\frac12\sum_{k=0}^{2n-1}\binom{2n-1}k\;.$$
Can you finish the proof from here?
Added: Alternatively, you can use the first identity to rewrite the desired result as
$$\sum_{k=0}^{n-1}\binom{2n-1}k=2^{2n-2}$$
and get there even quicker.
A: A different solution. The left hand side is the number of ways to pick any number of students between $0$ and $n$ and then picking one to be the "leader." To show this equal to the right hand side, note that we can pick a leader first, giving $2n$ choices, and then for the remaining $2n-1$ people, each can either be in the group or not, which gives $2^{2n-1}.$ 
The point now is that for the remaining $2n-1$ people, any group which takes $k\le n-1$ of them has a corresponding complement which takes $2n-1 - k \ge n$ of them, so exactly half of these groups have at most $n$ people, giving the desired $n\cdot 2^{2n-1}.$
A: To evaluate that sum, consider
$\sum_{k=0}n \binom{2n}{k} x^k.$ If This is half of the sum from $0$ to $2n,$ which is equal to $(1+x)^{2n}.$ To get the $k$ term,integrate the previous sum with respect to $x,$ and see what happens.
