Hints on solving $\sum_{i=1}^n i \cdot 2^{i-1}$

Question

I need hints on solving this summation.

Problem

$\sum_{i=1}^n i \cdot 2^{i-1} = \ldots$

Thanks in advance.

Have you heard the term Arithmetic-Geometric Progression/AGP? If not, read this article — john doe, Sep 23 '17 at 05:29
BTW you could find a few similar questions simply by putting your sum into Approach0. In this case - since it is a question which (with some minor variations) is asked very often - looking in the frequently asked question in the summation tag would be a reasonable thing to do, too. — Martin Sleziak, Sep 23 '17 at 06:08
Here are two more similar questions: How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$? and What is the sum of $\sum\limits_{i=1}^{n}ip^i$? — Martin Sleziak, Sep 23 '17 at 06:10

score 1 · Answer 1 · answered Sep 23 '17 at 05:33

1

Using a geometric series we define a polynomial by $$f(x)=\sum_{i=0}^nx^i=\frac{1-x^n}{1-x}$$ is the sum of the first n terms of a geometric progression. This is a standard result about geometric progression - if you are unfamiliar I believe others have added notes to guide you.

Now, our function $f$ is indeed differentiable as it is a finite polynomial.

So $$\begin{align*} \sum_{i=1}^nix^{i-1}&=\frac d{dx}\left(\sum^n_{i=0}x^i\right) \\&=\frac d{dx}\left(\frac{1-x^n}{1-x}\right) \end{align*}$$

paying close attention to the sum index. Can you take it from here?

answered Sep 23 '17 at 05:33

Jihoon Kang

949

What I get is this,
$\sum_{i=1}^n ix^{i-1} = \dfrac{nx^{n-1} (x - 1) - x^n + 1}{(x-1)^2} $.

Then, I substitute $x = 2$, and I get

$n\cdot 2^{n-1} - 2^n + 1$.

What is wrong with my work?
– Muhamad Abdul Rosid Sep 23 '17 at 05:40
@MuhamadAbdulRosid I think you are close to the answer - check your differentiation. Or it's possible you are correct - I'm afraid I cannot check right now but later – Jihoon Kang Sep 23 '17 at 05:49

score 1 · Answer 2 · answered Sep 23 '17 at 05:50

Let $S= \sum_{i=1}^{n} i.2^{i-1} = 1(1) + 2(2) + 3(4) + 4(8) + ... + n(2^{n-1}) $
$\quad\qquad\qquad 2S = \sum_{i=1}^{n} i.2^{i} = 1(2) + 2(4) + 3(8) + ... + (n-1)(2^{n-1}) + n(2^n)$

Therefore, $2S-S = -(1+2+4+8+... + 2^{n-1}) + n(2^n) = - (2^n-1) + n(2^n)=2^n(n-1) +1$

score 0 · Answer 3 · answered Sep 23 '17 at 05:40

0

$\sum_{i=1}^{n}i\times x^{i-1} = (\sum_{i=1}^{n}x^{i})' = (\frac{x^{n+1}-x}{x-1})' = \frac{\big((n+1)x^{n} - 1\big)(x-1) - (x^{n+1}-x)}{(x-1)^{2}} $

Let $x=2$. Then, $\sum_{i=1}^{n}i\times 2^{i-1} = \big((n+1)2^{n} - 1\big) - (2^{n+1}-2)$

answered Sep 23 '17 at 05:40

Hasan Heydari

1,187
1
8
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Hints on solving $\sum_{i=1}^n i \cdot 2^{i-1}$

3 Answers3