TITLE: Prove $\sum_{k= 0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}$ using the binomial theorem for $n\geq1$
QUESTION [5 upvotes]: 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.

REPLY [7 votes]: 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}
$
{"set_name": "stack_exchange", "score": 5, "question_id": 2550936}