Related Theorem of Binomial Theorem Proving that for any whole number $n$, the following identity holds: $$\sum^{n}_{i=1}{n\choose{i}}i=n\times2^{n-1}$$  So, I memorized this formula for preparing for math contests, but I think it's totally useless, and less efficient than actually understand what this identity means, rather than just brute force memorizing it.  I'm guessing that $n\times2^{n-1}$ can potentially mean that there're $n-1$ on-off switches, and the extra $n$ confuses me.  The sum on the left doesn't really even make sense to me.  Someone mind to explain this in detail to me?  Thanks!
 A: Method $1$:
From the binomial theorem, we have
$$\sum_{k=0}^n \dbinom{n}k x^k = (1+x)^n$$
Differentiate both sides to get
$$\sum_{k=0}^n k \dbinom{n}k x^{k-1} = n(1+x)^{n-1}$$
Set $x=1$, to get what you want. You could obtain related results by differentiating or integrating it multiple times.
Method $2$:
Another way to look at it is to realize that
$$i\dbinom{n}i = n \binom{n-1}{i-1}$$
Hence,
$$\sum_{k=1}^n k \dbinom{n}k x^{k-1} = n \sum_{k=1}^n \dbinom{n-1}{k-1} x^{k-1} = n \sum_{k=0}^{n-1} \dbinom{n-1}{k} x^{k} = n (1+x)^{n-1}$$
Method $3$: Combinatorial interpretation, which Danny Cheuk has already provided.
A: Hint:  Imagine you're choosing a committee of $i$ people from a total of $n$ people, but then one of those $i$ people gets to be the leader.
A: One way to prove it algebraically is to notice that
$$\binom{n}i=\frac{n!}{i!(n-i)!}\cdot  i=\frac{n!}{(i-1)!(n-k)!}=n\binom{n-1}{i-1}\;,\tag{1}$$
so that $$\sum_{i=1}^n\binom{n}ii=n\sum_{i=1}^n\binom{n-1}{i-1}=n\sum_{i=0}^{n-1}\binom{n-1}i=n2^{n-1}\;.$$ The identity $(1)$ is surprisingly useful.
A combinatorial proof is also possible. Start with the easy side: $n2^{n-1}$ is the way to choose one of $n$ people to be a team captain and then to choose any subset of the other $n-1$ people to fill out the team. $\binom{n}ii$ is the number of ways to pick a team of $i$ people from the original set of $n$ and then choose one of them to be the captain. Since a captain is required, the possible team sizes run from $1$ through $n$.
