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Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 24, 2011"

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{{:AoPSWiki:Problem of the Day/June 24, 2011}}
 
{{:AoPSWiki:Problem of the Day/June 24, 2011}}
 
==Solutions==
 
==Solutions==
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{{potd_solution}}
 
 
We see that <math>10=2\cdot5</math>, <math>40=5\cdot8</math>, <math>88=8\cdot11</math>, <math>154=11\cdot14</math>, and <math>238=14\cdot17</math>, so each term in the sum is of the form <math>\frac{1}{n(n+3)}=\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)</math>.
 
We see that <math>10=2\cdot5</math>, <math>40=5\cdot8</math>, <math>88=8\cdot11</math>, <math>154=11\cdot14</math>, and <math>238=14\cdot17</math>, so each term in the sum is of the form <math>\frac{1}{n(n+3)}=\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)</math>.
  
 
Therefore, the sum is
 
Therefore, the sum is
  
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<math>\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots=</math>
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<math>\frac{1}{3}\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots\right)=</math>
  
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<math>\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac15-\frac18\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\cdots</math>
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<math>\frac{1}{3}\left(\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac15-\frac18\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\cdots\right)</math>
  
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Eventually, all the fractions that occur later in the sum tend to <math>0</math> and all of them except for <math>\frac{1}{2}</math> cancel out, leaving <math>\frac{1}{2}</math>. -AwesomeToad
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Eventually, all the fractions that occur later in the sum tend to <math>0</math> and all of them except for <math>\frac{1}{2}</math> cancel out, leaving <math>\frac{1}{3}*\frac{1}{2}=\boxed{\frac{1}{6}}</math>.

Latest revision as of 10:07, 24 June 2011

Problem

AoPSWiki:Problem of the Day/June 24, 2011

Solutions

We see that $10=2\cdot5$, $40=5\cdot8$, $88=8\cdot11$, $154=11\cdot14$, and $238=14\cdot17$, so each term in the sum is of the form $\frac{1}{n(n+3)}=\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)$.

Therefore, the sum is

$\frac{1}{3}\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots\right)=$

$\frac{1}{3}\left(\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac15-\frac18\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\cdots\right)$

Eventually, all the fractions that occur later in the sum tend to $0$ and all of them except for $\frac{1}{2}$ cancel out, leaving $\frac{1}{3}*\frac{1}{2}=\boxed{\frac{1}{6}}$.

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