variation of the binomial theorem Why does: 
$$ \sum_{k=0}^{n} k \binom nk p^k (1-p)^{n-k} = np $$ ?
Taking the derivative of: 
$$ \sum_{k=0}^{n}  \binom nk p^k (1-p)^{n-k} = (1 + [1-P])^n = 1 $$ 
does not seem useful, since you would get zero. And induction hasn't yet worked for me, since – during the inductive step – I am unable to prove that: 
$$ \sum_{k=0}^{n+1} k \binom {n+1}{k} p^k (1-p)^{n+1-k} = (n+1)p  $$ 
assuming that: 
$$ \sum_{k=0}^{n} k \binom {n}{k} p^k (1-p)^{n-k} = np  $$ 

You can get rid of the $k$ by cancelling it with $k!$ from the binomial coefficient and simplifying like this: $$\begin{align} \sum_{k=0}^{n} k \binom nk p^k (1-p)^{n-k} &=  \sum_{k=1}^{n} k\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} =  n\sum_{k=1}^{n} \frac{(n-1)!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}\\ &=  n\sum_{k=1}^{n} \binom{n-1}{k-1} p^k (1-p)^{n-k}=n\sum_{k=0}^{n-1} \binom{n-1}{k} p^{k+1} (1-p)^{n-(k+1)}\\&={np\sum_{k=0}^{n-1} \binom{n-1}{k} p^{k} (1-p)^{n-1-k}}=np\cdot(p+1-p)^{n-1}=np\end{align}$$
