Having problems proving $ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (n+2)2^{n-1} $ I'm trying to prove the following
$$ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (n+2)2^{n-1} $$
using the identity $$ \sum_{r=0}^{n} \binom{n}{r} = 2^{n} $$
but I'm not able to. This is what I did so far, 
$$ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (1) \binom{n}{0} + (2) \binom{n}{1} + ... + (n+1) \binom{n}{n} $$

Starting with the binomials expansion
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} t^{r} = (1+t)^{n}
\end{align}
it can be seen by differentiation that
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} r t^{r} = n t (1+t)^{n-1} = n \left[ (1+t)^{n} - 
(1+t)^{n-1} \right].
\end{align}
Now by adding the original series to this result then
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} (r+1) t^{r} = (n+1) (1+t)^{n} - n (1+t)^{n-1}.
\end{align}
Now let $t=1$ to obtain
\begin{align}
\sum_{r=0}^{n} \binom{n}{r} (r+1) &= 2^{n} (n+1) - 2^{n-1} n \\
&= 2^{n-1} (n+2)  
\end{align} 
