AoPS Wiki talk:Problem of the Day/June 24, 2011

Revision as of 20:56, 23 June 2011 by AwesomeToad (talk | contribs)

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

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