Prove $\sum_{k= 0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}$ using the binomial theorem for $n\geq1$

Question

I do as follows

$(1+x)^n= \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$

differentiating in both sides

$n(1+x)^{n-1}= \binom{n}{1}+2\binom{n}{2}x+3\binom{n}{3}x^2...+n\binom{n}{n}x^{n-1}$

then, for $x=1$

$n \cdot2^{n-1} = \sum_{k=1}^{n}k\binom{n}{k}$

From here, I don't know how to proceed, I have tried to multiply both sides for $(1+x)$ to make $n \cdot2^{n}$ but then the summatory changes, I don't know how to prove the equality.

Answer 1

When you're trying to prove something and get stuck, always try a few examples to make sure it's actually true :)

$$\sum_{k= 0}^{n} k \binom{n}{k} {\LARGE \substack{? \\ =}} n \cdot 2^{n}$$

Here, let's try $n = 1$. We get $0 \binom{1}{0} + 1 \binom{1}{1} = 1$ on the left, and we get $1 \cdot 2^1 = 2$ on the right. So the statement is not correct.

The correct statement is $$\sum_{k= 0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}$$

which you are very close to proving, I can see.

Answer 2

A simpler way, not original, direct from the binomial theorem.

$\begin{array}\\ S &= \sum_{k=0}^n k \binom{n}{k}\\ &= \sum_{k=0}^n (n-k) \binom{n}{n-k} \qquad\text{replace } k \text{ by } n-k\\ &= \sum_{k=0}^n n\binom{n}{n-k}-\sum_{k=0}^n k\binom{n}{n-k}\\ &= n\sum_{k=0}^n \binom{n}{k}-\sum_{k=0}^n k\binom{n}{k} \quad\text{since } \binom{n}{n-k}=\binom{n}{k}\\ &= n2^n-S\\ \text{so}\\ S &=n2^{n-1}\\ \end{array} $