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

Problem

AoPSWiki:Problem of the Day/June 24, 2011

Solutions

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

$\frac{1}{3}\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\cdots\right)=$

$\frac{1}{3}\left(\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac15-\frac18\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\cdots\right)$

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 $\frac{1}{3}*\frac{1}{2}=\boxed{\frac{1}{6}}$ .

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

Problem

Solutions