Proof $\sum_{k=0}^{n}{k\binom{n}{k}}=n2^{n-1}$

Question

$\sum_{k=0}^{n}{k\binom{n}{k}}=n2^{n-1}$

$n2^{n-1} = \frac{n}{2}2^{n} = \frac{n}{2}(1+1)^n = \frac{n}{2}\sum_{k=0}^{n}{\binom{n}{k}}$

That's all I got so far, I don't know how to proceed

Answer 1

Two approaches:

First Approach: Consider $(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$, differentiate both sides with respect to $x$ and substitute $x=1$.

Second Approach: Use the identity $k\binom{n}{k}=n\binom{n-1}{k-1}$ for $k\ge 1$ and apply the result that $\sum_{k=0}^n \binom{n}{k}=2^n$

Answer 2

Hint. One has $$ k{\binom{n}{k}}=n{\binom{n-1}{k-1}},\quad n>0,k>0. $$

Answer 3

Take the function $(x+1)^n = 1^nx^0 {n \choose 0} \ + 1^{n-1}x^1{ n \choose 1} \ + 1^{n-2}x^2{ n \choose 2}+... + 1^{1}x^{n-1}{ n \choose {n-1}}+\ + 1^{0}x^n{ n \choose n} $

Differentiating both sides: $n(x+1)^{n-1} = \sum_{k=0}^{n}k{n \choose k}x^{k-1}$

Plugging $x=1$ on both sides gives us $n(1+1)^{n-1}=\sum_{k=0}^{n}k{n \choose k}=n2^{n-1}$

Answer 4

There is also a combinatorial proof. Imagine there is a set of $n$ elements and you want to choose a subset with one special element in it. (it can also be the only element of the subset). How can you do that? You can first choose the special element ($n$ options for that) and then choose a subset from the remaining $n-1$ elements. So the number of options is $n2^{n-1}$.

Another way is to find the number of options to choose a subset of a specific size $k$ and then sum on all possible values of $k$. So to choose a subset of size $k$ you need to choose $k$ elements outside of $n$ which is $\binom nk$ options and then you have $k$ options to choose the special element. So it is $k\binom nk$ options to choose a subset of size $k$. Sum on all possible values of $k$ and you will get the required identity.

Answer 5

I want to give another way of doing it, but all the other answers are very good (I really like the differentiation one). Mine is going to be more of simple arithmetic and change of indices.

Note that the binomial coefficient can be written as $${n\choose k} = \frac{n!}{k!(n-k)!}$$ and that $$\frac{k}{k!} = \frac{k}{k(k-1)!} = \frac{1}{(k-1)!}$$ If we substitute back into the sum $$\sum_{k=0}^n k{n\choose k} = \sum_{k=0}^n\frac{k}{k!}\frac{n!}{(n-k)!} = \sum_{k=0}^n\frac{n!}{(k-1)!(n-k)!}$$ by making a new index $m =k-1$, what we get in the sum is $$\sum_{m=0}^{n-1}\frac{n!}{m!(n-(m+1))!} = \sum_{m=0}^{n-1}n\frac{(n-1)!}{m!((n-1)-m)!}$$

Using now the binomial theorem, that states $$(a+b)^n=\sum_{m=0}^{n}{n\choose m}a^{n-m}b^m$$ setting $a=b=1$, we get $$2^n = \sum_{m=0}^{n}{n\choose m}$$

That's pretty similar to what we got earlier, setting $j=n-1$ $$\sum_{m=0}^{n-1}n\frac{(n-1)!}{m!((n-1)-m)!} = n\sum_{m=0}^j\frac{j!}{m!(j-m)!} = n\sum_{m=0}^j{j\choose m} = n2^j = n2^{n-1}$$