Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Lately, I was wondering if there exists a closed expression for $2^0+2^1+\cdots+2^n$ for any $n$?

share|improve this question
6  
3  
Did you try a few values of $n$? –  lhf Nov 29 '11 at 21:22

4 Answers 4

Another nice way to see this is you can write the number in base 2 : The sum $$ 2^0 + 2^1 + \dots + 2^n = (011\dots1)_2 = (100\dots0)_2 - (0\dots 01)_2 = 2^{n+1} - 1. $$ where the writing of that number in base 2 has $n+1$ digits.

Hope that helps,

share|improve this answer
    
This is the cleverest of the lot. :) –  J. M. Nov 30 '11 at 3:20
    
Wow. And I was wondering if it was even worth posting. XD Hahahaha –  Patrick Da Silva Nov 30 '11 at 4:30

As the sum of a geometric sequence, $1+2^1+2^2+\ldots+2^n = \frac{2^{n+1}-1}{2-1}=2^{n+1}-1$.

You can see it by computing $(1+2^1+2^2+\ldots+2^n)(2-1)$ and distribute. All the terms except for $2 \cdot 2^n - 1$ will cancel each other out.

share|improve this answer

that is geometric series

$1+q+...+q^n=\frac{1-q^{n+1}}{1-q}$

with quotient q=2

$2^{0}+2^{1}+...+2^{n}=\frac{1-2^{n+1}}{1-2}=2^{n+1}-1$

share|improve this answer

Clue:

If you are searching closed formulas to a series like:

$a+ab+ab^2+ab^3+\ldots+ab^n = S_n$

You can use the follow trick:

$a+ab+ab^2+ab^3+\ldots+ab^n = S_n$

so,

$b(a+ab+ab^2+ab^3+\ldots+ab^n) = bS_n = ab+ab^2+ab^3+ab^4+\ldots+ab^n+ab^{n+1}=$

$a+ab+ab^2+ab^3+\ldots+ab^n -a +ab^{n+1}= bS_n \iff$

$S_n-a+ab^{n+1}=bS_n \iff S_n(b-1)=a(b^{n+1}-b) \iff$

$$S_n=\frac{a(b^{n+1}-b)}{b-1}$$

But, $b=1$ isn't covered by this formula, but if b=1, the series become

$S_n=a+a+a+...+a$ where $a$ appears $n$ times and $S_n=na$.

To your question,

$a=1=2^0$ and $b=2$,

so

$S_n=a+ab+ab^2+ab^3+\ldots+ab^n=2^0+2^1+\ldots+2^n=\frac{1(2^{n+1}-1)}{2-1}=2^{n+1}-1$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.