Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 24, 2011"
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==Solutions== | ==Solutions== | ||
− | {{ | + | 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}{3}\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots\right)=</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}{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 , , , , 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 .