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Given the following sum formula:

$\sum\limits_{i=1}^{n} (i\cdot 2^{n-i}) $

Can you help me out to a simplify the formula and provide an formula without a Sigma sign?

I know that I cannot just split the Sigma like that:

$\sum\limits_{i=1}^{n} (i\cdot 2^{n-i}) \ne (\sum\limits_{i=1}^{n} i) \cdot (\sum\limits_{i=1}^{n}2^{n-i}) $

I am wondering if there is something else I can do

Stasel
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  • Did you write it correctly? As written it is just $2^{n-1}\times \sum_{i=1}^ni$ – lulu Aug 12 '18 at 14:29
  • @lulu sorry, there was a mistake in the question. I have edited the question. Thanks for the heads up – Stasel Aug 12 '18 at 14:30
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    Expressions of this sort can be analyzed via geometric series. For an infinite version, note that $f(x)=\frac {1}{1-x}=\sum_{i=0}^{\infty} x^i$ and differentiation yields $f'(x)=\sum_{i=1}^{\infty} ix^{i-1}$ which is akin to yours (you'll need to work with partial sums for the geometric series, of course). – lulu Aug 12 '18 at 14:34
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    Already asked zillion times. –  Aug 12 '18 at 14:36
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    Let the sum be $S$. Calculate $S = 2S - S$ by rearranging indices. – xbh Aug 12 '18 at 14:40
  • This is not a sum of $n$ products. It is a sum of products where the first term consists of $n$ products. – Ahmad Bazzi Aug 12 '18 at 14:47

2 Answers2

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Hint:

$$1\cdot2^{n-1}+2\cdot2^{n-2}+\cdots n\cdot2^0 \\=0\cdot2^{n-1}+1\cdot2^{n-2}+\cdots (n-1)\cdot2^0+2^{n-1}+2^{n-2}+\cdots2^0 \\=\frac{1\cdot2^{n-1}+2\cdot2^{n-2}+\cdots n\cdot2^0}2-\frac{n\cdot2^0}2+2^n-1.$$

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Following xbh's hint: $$S=\sum\limits_{i=1}^{n} (i\cdot 2^{n-i})=\color{red}{1\cdot 2^{n-1}}+2\cdot 2^{n-2}+\cdots+\color{blue}{(n-1)\cdot 2^1}+n\cdot 2^0\\ 2S=2\sum\limits_{i=1}^{n} (i\cdot 2^{n-i})=1\cdot 2^{n}+\color{red}{2\cdot 2^{n-1}}+\cdots+(n-1)\cdot 2^2+\color{blue}{n\cdot 2^1}.$$ Subtract $2S-S$: $$S=2^n+\color{red}{2^{n-1}}+\cdots+\color{blue}{2}-n=\frac{2(2^n-1)}{2-1}-n=2^{n+1}-2-n. $$

farruhota
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