Find the value of $\sum_{k=1}^{n} k \binom {n} {k}$

Question

I was assigned the following problem: find the value of $$\sum_{k=1}^{n} k \binom {n} {k}$$ by using the derivative of $(1+x)^n$, but I'm basically clueless. Can anyone give me a hint?

Answer 1

Note that $$(1+x)^n= \sum_{k=0}^{n} \binom {n} {k} x^k$$

$$f'(x)=n \cdot (1+x)^{n-1}= \sum_{k=1}^{n} k \binom {n} {k} x^{k-1}$$

Whence, we have:

$$\sum_{k=1}^{n} k \binom {n} {k} = f'(1) = n \cdot 2^{n-1}$$

Answer 2

Notice that $\displaystyle S = \sum_{k=0}^{n} k \binom {n} {k} = \sum_{k=0}^{n}(n-k)\binom {n} {n-k} = n\sum_{k=0}^{n}\binom {n} {n-k}-S$

so, as $\displaystyle \binom {n} {n-k}=\binom {n} {k}$ we have $\displaystyle 2S = n\sum_{k=0}^{n}\binom {n} {k} = n2^n$ and so $\displaystyle S = n2^{n-1}.$

Answer 3

Imagine tossing a fair coin $n$ times. Then the mean number of heads is $$\sum_0^n k\binom{n}{k}\frac{1}{2^n}.\tag{1}$$ We compute the mean another way. Let random variable $X_i$ be $1$ if we get a head on the $i$-th toss, and $0$ otherwise. Then the number $Y$ of heads is given by $$Y=X_1+X_2+\cdots +X_n,$$ and therefore by the linearity of expectation we have $$E(Y)=E(X_1)+E(X_2)+\cdots +E(X_n).$$ Since $X_i=1$ with probability $\frac{1}{2}$, we have $$E(Y)=\frac{n}{2}.\tag{2}$$ Now comparison of (1) and (2) gives the result.

Remark: This is an example of a Mean Proof.

Answer 4

Recall that $$ (1+x)^n = \sum_{k=0}^{n} x^k \binom {n} {k} $$ and thus $$ ((1+x)^n)' = \sum_{k=1}^{n} k x^{k-1} \binom {n} {k} $$ Now, calculate the left-hand side, and then think which value of $x$ could be a good choice.

Answer 5

Hint : set $$f(x)=(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$$ What you get when you evaluate $f'(1)$ ?

Note that you sum can begin from $0$ since the term with $k=0$ is $0$.