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: 
$$ \sum_{k=0}^{n} k \binom {n}{k} p^k (1-p)^{n-k} = np  $$ 

This is saying that the average number of coin flips if you flip $n$ fair coins each with heads probability $p$ is $np$ (you are calculating the mean of a Binomial(n,p) distribution). 
You can use linearity of expectation - if $Y = \sum_i X_i$ where $X_i$ is Bernoulli(p) and there are $n$ such $X_i$, then $Y$ is Binomial(n,p) and has that distribution. Then, $E[Y] = \sum_i E[X_i]$ and you can easily show $E[X_i]=p$. 
If you want to proceed in this manner, note $\binom{n}{k}=\frac{n}{k} \binom{n-1}{k-1}$. You can find the full details of the proof on Proof Wiki.
