Prove the formula $\sum_{k=1}^n k\binom{n}{k} = n \cdot 2^{n-1}$ for all integers $n > 0$ I just got to this question and I became a question mark. I wonder if anyone can help me with this one, because I don't even know how to begin to tackle this problem.
The question:

Prove the formula $\sum_{k=1}^n k\binom{n}{k} = n \cdot 2^{n-1}$ for all
  integers $n > 0$. We accept a proof based on manipulations of formulas
  and do not demand a proof by mathematical induction here.

Hint: Write out the first few terms in the sum $\sum_{k=1}^n k\binom{n}{k}$ and factor out $n$. Then use the formula $\sum_{j=0}^m \binom{m}{j}  = 2^m$ that is valid for all non-negative integers $m$, particularly $m = n−1$.
Please help. I don't understand it.
 A: Need to follow hint your professor said, calculate a few terms, and factor out n:
$$k\binom nk = k \frac{n!}{k!(n-k)!} = n\frac{(n-1)!}{(k-1)!(n-k)!} = n\frac{(n-1)!}{(k-1)!((n-1) - (k - 1))!} = n\binom {n-1}{k-1}$$
Follow the rest hints to finish the proof.
Now we use the hint
$$\sum_{j=0}^m \binom{m}{j}  = 2^m$$
$$\sum_{k=1}^n k \frac{n!}{k!(n-k)!} = \sum_{k=1}^n n\binom {n-1}{k-1}$$
$$=n\sum_{k=1}^n\binom {n-1}{k-1}$$
let $ j = k -1$, so $ j = 1 ... (n-1)$, using hint with $m = n -1$
$$=n\sum_{j=0}^{n-1}\binom {n-1}{j} = n 2^{n-1}$$
A: I will give an inductive proof using the fact that $ \sum_{k=0}^n \tbinom n k = 2^n$ as well as Pascal's identity. 
The base case is straightforward.  
Suppose it holds for some $n$. Then $$\sum_{k=1}^{n+1} k\binom {n+1} k = n+1 + \sum_{k=1}^{n} k\binom {n} {k-1} + \sum_{k=1}^{n} k\binom {n}{k} = n + 1 + \sum_{k=1}^{n} (k-1)\binom {n} {k-1} + \sum_{k=1}^{n} \binom {n} {k-1} + \sum_{k=1}^{n} k\binom {n}{k} = (n+1) + (n \cdot 2^{n-1} - n) + (2^{n} - 1) + (n \cdot 2^{n-1}) = n\cdot2^n + 2^n = (n+1)2^{(n+1)-1}$$. 
hence it holds for $n+1$. 
This completes the proof. 
A: Hint:
Can you calculate $$\sum_{k=1}^nk\binom nk x^{k-1}?$$
A: I a giving a combinatorial proof.
Consider a set $S$ with $n$ elements. Now, calculate the number of elements in every subsets of this set $S$. So, in the subsets containing $k$-elements, there are $\color{blue}{k\binom nk}$ elements in total. So, for $0\le k\le n$, the number of all the elements in all the possible subsets is $$\color{red}{{\sum\limits_{k=1}^nk\binom nk}}\tag 1.$$ Now, count in a different way. Count the total number of elements of a subset along with its complementary subset. 
Say $A$ be a subset of $S$, then #$A+$#$A^c=\color{blue}{n}$. And earlier, there was $2^n$ subsets. So, while counting the number of subsets along with their complements, we should have $2^n/2=\color{blue}{2^{n-1}}$ subsets to count, each containing $n$ elements.
So, in this way, you get, the number of all the elements in all the possible subsets is $$\color{red}{n\times 2^{n-1}}.\tag 2$$
In both cases, we counted same quantity, so, the results should be equal. So, from $(1)$ and $(2)$ we get, $$\sum k\binom nk=n\times 2^{n-1}.$$
