find a formula for $\sum_{k=1}^{n} k {n \choose k}$ find a formula for $\sum_{k=1}^{n} k {n \choose k}$ and prove it
I know that $k \cdot {n \choose k}=\frac{n!}{\left( n-k\right)! \cdot \left( k-1\right)!}=n \cdot \frac{\left( n-1\right)! }{\left( n-k\right)! \cdot \left( k-1\right)! }=n \cdot {n-1 \choose k-1}$
and I tried to do something with it but i failed.
 A: Note that $\binom{n}{k}$ counts the number of ways to choose $k$ elements from an $n$-element set; having done this, there are $k$ ways to "mark" a special element from those $k$.
In other words, $k\binom{n}{k}$ is the number of ways to choose a subcommittee of size $k$ from a group of $n$ people, and name one of the $k$ subcommittee members as chair.
Summing over these, you are looking at all of the possible ways to choose a non-empty subcommittee from $n$ people, and name a chair.
But, we can count this another way: by first choosing a chairman in one of $n$ ways, then choosing any subset (empty or non-empty) of the remaining $n-1$ people; there are $2^{n-1}$ ways to do this.  So, it should be the case that
$$
\sum_{k=1}^{n}k\binom{n}{k}=n2^{n-1}.
$$
A: $$\sum_{1\le k\le n}k\binom nk=n\sum_{1\le k\le n}\binom{n-1}{k-1}=n\sum_{0\le r\le n-1}\binom{n-1}r=n(1+1)^{n-1}$$
Using Binomial Theorem formula  $$(a+b)^{n-1}=\sum_{0\le r\le n-1}\binom{n-1}r a^r b^{n-1-r}$$ for positive integer index, putting $a=b=1$ 
A: $(1+x)^n=C_0 + C_1 x+ C_2x^2+\dots C_n x^n$
If you can differentite this with $x$, you would get...
$n(1+x)^{n-1}=C_1 +2.C_2x+\dots+n.C_nx^{n-1}$
Substitute $1$ for $x$ you would get :
$n2^{n-1}=C_1 +2.C_2+\dots +n.C_n$
P.S : Here $C_i= \binom{n}{i}$
A: $$
\sum_{k=1}^{n}k\binom{n}{k}=\sum_{k=1}^{n}k\frac{n}{k}\binom{n-1}{k-1}=\sum_{k=1}^{n}n\binom{n-1}{k-1}=n\sum_{k=0}^{n-1}\binom{n-1}{k}=n2^{n-1}.
$$
