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

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<cmath>1+\frac{1}{2}+\frac{1}{4}+\cdots</cmath>
 
<cmath>1+\frac{1}{2}+\frac{1}{4}+\cdots</cmath>
 
is a geometric series, which famously converges to 2.  This can by applying the formula for infinite geometric sums, <math> \frac{a_1}{1-r} </math> where <math> a_1 </math> is the first term and <math> r </math> is the ratio. For this series, the sum is <math> \frac{1}{1-\frac{1}{2}}=2 </math>. Thus, since the problem asks for twice this number, the answer is <math>\boxed{4}</math>.
 
is a geometric series, which famously converges to 2.  This can by applying the formula for infinite geometric sums, <math> \frac{a_1}{1-r} </math> where <math> a_1 </math> is the first term and <math> r </math> is the ratio. For this series, the sum is <math> \frac{1}{1-\frac{1}{2}}=2 </math>. Thus, since the problem asks for twice this number, the answer is <math>\boxed{4}</math>.
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<noinclude>[[category:Problem of the Day]]</noinclude>

Revision as of 08:26, 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 $\boxed{4}$.