Counting arguments Given one prove the other identity Given:
$${n \choose 0} + {n \choose 1} + {n \choose 2} + \cdots + {n \choose n} = 2^n$$ 
Prove the following in 2 ways.
$$ {n \choose 1} + 2 {n \choose 2} + 3 {n \choose 3} + \cdots + n{n\choose n} = n 2^{n-1}.$$
I have already figured out one: 
$$
\sum_0^n k\binom{n}{k}
$$
Makes up part of the computation for the expected size of a group if you randomly select items from the total.  This expected value is half of the size of the total.  Thus:
$$
\frac{\sum_0^n k\binom{n}{k}}{\sum_0^n \binom{n}{k}} = \frac{n}{2}
$$
With the result that $\sum_0^n\binom{n}{k} = 2^n$, this gives:
$$
\sum_0^n k\binom{n}{k} = \frac{n}{2}2^n = n2^{n - 1}
$$
Which then equals the RHS
But I cannot figure out another way. Thanks for any help
 A: Second way can be done by blowing apart the binomials: $$\frac{k}{n}{n\choose k}=\frac{k}{n}\frac{n!}{(n-k)!k!}=\frac{(n-1)!}{(n-k)!(k-1)!}={n-1\choose k-1}$$
Now sum those up from $k=1$ to $k=n$ and you will get $2^{n-1}$.
Differentiating the binomial theorem is the "first way".  OP's method is actually a third way, a nice combinatorial proof.
A: $$
\sum_{k=0}^n k\binom{n}{k} = \sum_{k=1}^n \frac{k n!}{k!(n-k)!} \\= \sum_1^n \frac{n!}{(k-1)!(n-k)!} \\
=n\sum_1^n \frac{(n-1)!}{(k-1)!(n-k)!} \\
=n\sum_0^{n-1} \frac{(n-1)!}{k!(n-1-k)!} \\
=n2^{n-1}
$$
A: We are going to count the number of ways of making a team of any size less or equal to $n$ and then choose one as the leader of the team.
If the team is of size $k$ then the number of ways is $k{n\choose k}$. If the size of the team is not determined then we have $\sum_{k=1}^n k{n\choose k}$ ways.
The other counting is by first selecting the leader. Since there are $n$ people there are $n$  ways to select the leader. Then among the remaining $n-1$ people we can choose any subset of those $n-1$ people in $2^{n-1}$ ways to make a team. So in total we have $n\cdot 2^{n-1}$ ways.
A: Hint what's the derivative of original equation. Using derivatives can be one of the second way.
A: $$
(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k \\
n(1+x)^{n-1} = \sum_{k=1}^n k \binom{n}{k}x^{k -1}
$$
set $x=1$
