A finite binomial sum Is their an exact expression for the following sequence involving binomial coefficients $$\sum_{i=0}^n i\binom{n}{i}?$$
 A: One way to look at this is as follows.  Toss a coin $n$ times.  All $2^n$ outcomes are equally probable.  The number of ways to get $i$ heads out of $n$ is $\binom n i$, so the probability that that happens is $\binom n i /2^n$.
The expected number of heads is $n/2$, but the expected number of heads is
$$
\sum_{i=0}^n i\cdot\Pr(\text{exactly $i$ heads}) = \frac 1 {2^n}\sum_{i=0}^n i\binom n i.
$$
So
$$
\frac 1 {2^n}\sum_{i=0}^n i\binom n i = \frac n 2,
$$
and so
$$
\sum_{i=0}^n i\binom n i = n 2^{n-1}.
$$
A: You can use that $i \cdot \binom{n}{i} = n \cdot \binom{n-1}{i-1}$ for $i > 0$ and get
$$\sum_{i=0}^n i \cdot \binom{n}{i} = \sum_{i=1}^n i \cdot \binom{n}{i} = \sum_{i=1}^n n \cdot \binom{n-1}{i-1} = n \cdot \sum_{i=0}^{n-1} \binom{n-1}{i} = n \cdot 2^{n-1}.$$
A: You have $n$ children. How many ways are there to choose some of them to play tag, and then make one of the players "it"?
The number of kids playing tag will be some number, $i$. For each $i$, there are $\binom{n}i$ ways to choose the kids playing tag, then $i$ ways to choose which one is "it". Summing over $i$, the desired answer is $\color{red}{\sum_{i\ge0}i\binom{n}i}$.
On the other hand, you could first choose who will be "it", then choose who else is playing. There are $n$ ways to choose who is "it", then $2^{n-1}$ ways to choose a subset of everyone else to form the other players, so there are $\color{red}{n2^{n-1}}$ ways.
Since these count the same thing, they are equal!
A: We have: $\displaystyle \sum_{i=0}^n \binom{n}{i}x^i = (1+x)^n \to \displaystyle \sum_{i=0}^n i\binom{n}{i}x^{i-1} = n(1+x)^{n-1}$. Now put $x = 1$ to get $n2^{n-1} = \displaystyle \sum_{i=0}^n i\binom{n}{i}$.
