$$\sum_{i=1}^k i\cdot2^i$$
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?
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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}
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No, you can't rewrite $$\sum_{i=1}^ki\cdot2^i$$ as $$\sum_{i=1}^ki\cdot\sum_{i=1}^k2^i.$$
In less time that it took to post the question, you could have checked the case $k=2$ on a calculator: is $$1\cdot2+2\cdot4=(1+2)(2+4)?$$
However, you can write $$\sum_{i=1}^ki\cdot2^i=\sum_{i=1}^k2^i+\sum_{i=2}^k2^i+\sum_{i=3}^k2^i+\cdots+\sum_{i=k}^k2^i.$$ Right? Now can you take it from there?
Let $S$=$\sum_{i=1}^k i\cdot2^i$ Consider $2S$=$\sum_{i=1}^k i\cdot2^{i+1}$. Do $S-2S$ and the result is an G.P - an extra term $k2^{K+1}$.Find S.