Simplify a sum of n products

Question

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

Answer 1

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 $2S-S$: $$S=2^n+\color{red}{2^{n-1}}+\cdots+\color{blue}{2}-n=\frac{2(2^n-1)}{2-1}-n=2^{n+1}-2-n. $$

Answer 2

Hint:

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