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

(new problem talk page)
 
Line 3: Line 3:
 
==Solutions==
 
==Solutions==
 
{{potd_solution}}
 
{{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>.
 +
 +
Therefore, the sum is
 +
 +
<math>\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots=</math>
 +
 +
<math>\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac15-\frac18\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\cdots</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>. -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 $10=2\cdot5$, $40=5\cdot8$, $88=8\cdot11$, $154=11\cdot14$, 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

$\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots=$

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

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 $\frac{1}{2}$. -AwesomeToad

Invalid username
Login to AoPS