Prove $a_n =\sum_{i=0}^{n-1} 2^{i} (n-i)=2^{n+1}-n-2$

Question

How would one go about proving $\sum_{i=0}^{n-1} 2^{i} (n-i)=2^{n+1}-n-2$ while only manipulating the left side of the equation? I know this can be proven by induction, but I don't know how someone would get the equation $2^{n+1}-n-2$ from the sum $\sum_{i=0}^{n-1} 2^{i} (n-i)$. Can someone show me the steps to get from the summation to the equation? I would prefer the explanation to be in sigma notation if possible.

Answer 1

Hint: $a_{n+1}=\sum_{i=0}^n 2^i(n+1-i) = \sum_{i=0}^n 2^i(n-i) + \sum_{i=0}^n 2^i = a_n + 2^{n+1}-1\,$. Therefore $a_{n+1}-a_n = 2^{n+1}-1\,$, then add up and telescope.

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[ EDIT ] To answer the question asked in a comment, below is how telescoping works step by step.

$$ \require{cancel} (a_{n+1}-\cancel{a_n})+(\cancel{a_{n}}-\bcancel{a_{n-1}})+(\bcancel{a_{n-1}}-\cancel{a_{n-2}})+\cdots+(\bcancel{a_2}-a_1) \\ = (2^{n+1}-1)+(2^{n}-1)+(2^{n-1}-1)+\cdots+(2^2-1) $$

$$ a_{n+1} - a_1=(2^{n+1}+2^n+2^{n-1}+\cdots+2^2)-n $$

$$ a_{n+1} = a_1 + 2^2(2^{n}-1)-n=1+2^{n+2}-4-n=2^{n+2}-n-3 $$

Substituting $n+1 \mapsto n$ gives $a_n = 2^{n+1}-n-2\,$.

I would prefer the explanation to be in sigma notation if possible.

In sigma notation, the telescoping on the LHS can be written as:

$$ \begin{align} \sum_{i=1}^n (a_{i+1}-a_i) & =\sum_{i=1}^n a_{i+1}-\sum_{i=1}^n a_i=\sum_{i=2}^{n+1} a_{i}-\sum_{i=1}^n a_i \\ & = (\cancel{\sum_{i=2}^{n} a_i} + a_{n+1})-(a_1+\cancel{\sum_{i=2}^{n} a_i})=a_{n+1}-a_1 \end{align} $$

Answer 2

hint: $\displaystyle \sum_{j} 2^j(n-j)=\displaystyle n\sum_{j}2^j-\displaystyle \sum_{j}j\cdot2^j$. The latter sum can be found by using $x\dfrac{d}{dx}\displaystyle \sum_{j} x^j$, and evaluate the derivative at $x = 2$, and this is quite common.