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

Solution 1:

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} $

Solution 2:

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.