Difference between revisions of "AoPS Wiki talk:Problem of the Day/September 5, 2011"

 
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==Problem==
 
==Problem==
Simplify <cmath> 2(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\fac{1}{32}+\frac{1}{64}+\frac{1}{128}+\cdots) </cmath>  
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Simplify <cmath> 2(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\cdots) </cmath>  
  
 
==Solution==
 
==Solution==

Latest revision as of 08:28, 7 September 2011

Problem

Simplify \[2(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\cdots)\]

Solution

The series \[1+\frac{1}{2}+\frac{1}{4}+\cdots\] is a geometric series, which famously converges to 2. This can by applying the formula for infinite geometric sums, $\frac{a_1}{1-r}$ where $a_1$ is the first term and $r$ is the ratio. For this series, the sum is $\frac{1}{1-\frac{1}{2}}=2$. Thus, since the problem asks for twice this number, the answer is $\boxed{4}$.