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

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+ | 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>. | ||

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+ | 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>\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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+ | 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 |

## Revision as of 20:56, 23 June 2011

## Problem

AoPSWiki:Problem of the Day/June 24, 2011

## Solutions

*This Problem of the Day needs a solution. If you have a solution for it, please help us out by adding it.*
We see that , , , , and , so each term in the sum is of the form .

Therefore, the sum is

Eventually, all the fractions that occur later in the sum tend to and all of them except for cancel out, leaving . -AwesomeToad