Sum of series proof $\sum_{i=1}^{n} i2^i$

Question

How can I find the sum of $\sum_{i=1}^{n} i2^i$ in a closed form. I have a feeling I can do this using differentiation of an existing series but cant seem to find it.

Answer 1

$$ f(x) = \sum_{n=1}^k x^n = \frac{x^{k+1} - x }{x-1} $$

$$ \Rightarrow f'(x) = ??? \Rightarrow 2f'(2) = ??? $$

:-) not a complete answer to leave a chance for you to continue

Answer 2

You can do this without calculus. It is an arithmetico-geometric progression, i.e. a series in which each term is a product of a term in an arithmetic progression and the corresponding term in a geometric progression. The technique is same as that used to evaluate a geometric progression. Let $S = \sum\limits_{i=1}^{n} i2^i$. The "geometric component" of the series is $2,2^2,2^3,\cdots$ so the common ratio is 2. Therefore $$ \begin{align} 2S &= \sum_{i=1}^{n}i2^{i+1}=\sum_{i=1}^{n}(i+1-1)2^{i+1}\\ &= \sum_{i=2}^{n+1}i2^{i}-\sum_{i=1}^{n}2^{i+1}\\ &= S+ (n+1)2^{n+1}-(1)2^1 -\sum_{i=1}^{n}2^{i+1} \end{align}$$ Therefore $$S=(n+1)2^{n+1}-(1)2^1 -\sum_{i=1}^{n}2^{i+1}$$ which you can evaluate.

Answer 3

Similar to another answer, if you let $S_n = \sum_{i=1}^n i 2^i$, then $S_{n+1} = 2S_n + \sum_{i=1}^{n} 2^i = 2S_n + (2^{n+1} - 1)$. You can evaluate the first few terms with symbolic expressions and series and then prove a formula by induction using the recursion relation.

Answer 4

\begin{align} \sum_{i = 1}^{n}p^{i} &= {p^{n + 1} - p \over p - 1} \\[3mm] \sum_{i = 1}^{n}ip^{i - 1} &= {\left(n + 1\right)p^{n} - 1 \over p - 1} - {p^{n + 1} - p \over \left(p - 1\right)^{2}} \\[3mm] \sum_{i = 1}^{n}ip^{i} &= {\left(n + 1\right)p^{n + 1} - p \over p - 1} - {p^{n + 2} - p^{2} \over \left(p - 1\right)^{2}} \\[3mm] \mbox{Set}\ p = 2\,;\quad \sum_{i = 1}^{n}i\,2^{i} &= \left\lbrack\left(n + 1\right)2^{n + 1} - 2\right\rbrack - \left\lbrack2^{n + 2} - 4\right\rbrack \end{align}

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \sum_{i = 1}^{n}i\,2^{i} = \left(n - 1\right)2^{n + 1} + 2\quad} \\ \\ \hline \end{array} $$