Sum of combination My problem is:
Evaluate:
$$\sum_{i=0}^n i{n \choose i}$$
I only know that $$\sum_{i=0}^n{n \choose i} = 2^n$$ not so sure when an "i" is added.
What is the step of this evaluation?
 A: Because this is a statistics site, this question needs a statistical answer.  Here is one of many possible statistical approaches.
Draw a picture to plot $\binom{n}{i}$ against $i$ for all valid values $0\le i \le n$:

(Produced in R with the command n <- 8; barplot(choose(n, 0:n), names.arg=0:n).)
Upon dividing by the sum of these values (which the question correctly asserts is $2^n$) we obtain a discrete probability distribution $F$ for which
$${\Pr}_F(i) = 2^{-n}\binom{n}{i}.$$
The sum in question is by definition a multiple of the expectation of any random variable $X$ having this distribution:
$$\mathbb{E}_F(X) = \sum_i i{\Pr}_F(i) = \sum_{i=1}^{n} 2^{-n} \binom{n}{i}$$
whence
$$\sum_{i=1}^{n} i\binom{n}{i} = 2^n \mathbb{E}_F(X).$$
The plot strongly suggests this distribution is symmetric.  That reminds us of the fundamental identity
$$\binom{n}{i} = \binom{n}{n-i},$$
proving the symmetry.  Since the expectation of a symmetric distribution is in its middle, we immediately conclude that
$$\mathbb{E}_F(X) = (n+0)/2 = n/2.$$
The answer is now easy to obtain.

Those a tiny bit more advanced in probability will recognize the distribution $F$ as a Binomial distribution and also will know that a Binomial distribution is the sum of $n$ independent Bernoulli$(1/2)$ distributions.  Thus, upon recognizing the sum as a multiple of the expectation, they will immediately conclude that expectation is $n$ times the expectation of Bernoulli$(1/2)$, which is obviously $1/2$, immediately obtaining the sum as the product of $n$, $1/2$, and $2^n$.  (This solution did not invoke any symmetry argument.)
A: First of all, note that you can change the range of the dummy index $i$, because
$$
  \sum_{i=0}^n i {n \choose i} = \sum_{i=1}^n i {n \choose i}  \, , \qquad (*)
$$
since the first term of the original sum is equal to zero. Now, suppose that we have the following problem: we have $n$ people, and we want to select groups of people of any possible size choosing for each of the groups one group leader. In how many ways can we do this?
The strategy is to solve the problem in two different ways: the first way gives as a solution $(*)$, and the second way gives the desired summation. 
First way: for a group of size $i$, one way to make the selection it is to first choose the group from the $n$ available people, for which we have ${n\choose i}$ possibilities, and subsequently choose one of the $i$ people in the group to be the leader, which gives us $i{n\choose i}$ possibilities. Summing for each possible group size $i=1,\dots,n$,  the answer to the problem is exactly the desired sum $(*)$.
Second way: for a group of size $i$, we first choose one of the $n$ people to be the group leader, and subsequently choose $i-1$ people from the remaining $n-1$ to complete the group, giving us $n{n-1\choose i-1}$ possibilities. Summing for each possible group size $i=1,\dots,n$, the answer is
$$
  \sum_{i=1}^n n {n-1\choose i-1} = n \sum_{i=0}^{n-1} {n-1\choose i} = n\,2^{n-1} \, ,
$$
in which we adjusted the range of the dummy index $i$ and used the other identity you provided in your question, $\sum_{i=0}^n {n\choose i}=2^n$, which also has a simple combinatorial interpretation. Therefore,
$$
  \sum_{i=0}^n i {n \choose i} = n\,2^{n-1} \, .
$$
