Solving this summation: $\sum_{i=1}^k i\cdot2^i$ $$\sum_{i=1}^k i\cdot2^i$$
I'm working on a recurrence relation question and now I'm stuck at this point. I have no idea how to simplify this down to something I can work with.  Can I seperate the terms into
$$\sum_{i=1}^k i \cdot \sum_{i=1}^k 2^i$$ and then just use the geometric series? 

Consider the series
\begin{align}
\sum_{k=0}^{n} t^{k} = \frac{1-t^{n+1}}{1-t}.
\end{align}
Differentiate both sides with respect to $t$ to obtain
\begin{align}
\sum_{k=0}^{n} k t^{k-1} &= \frac{1}{(1-t)^{2}} \left( -(n+1) (1-t) t^{n}+(1-t^{n+1}) \right) \\
&= \frac{1 -(n+1) t^{n} + n t^{n+1}}{(1-t)^{2}}.
\end{align}
Now let $t=2$ to obtain
\begin{align}
\sum_{k=0}^{n} k \cdot 2^{k} = 2 -(n+1) 2^{n+1} + 2^{n+2} n = 2 + 2^{n+1}(n-1).
\end{align}
