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>. | 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 |
<math>\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots=</math> | <math>\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots=</math> |
Revision as of 23:08, 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