Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 24, 2011"
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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}{3}\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots\right)=</math> |
− | <math>\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac15-\frac18\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\cdots</math> | + | <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> |
− | 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>. | + | 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 .