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


AoPSWiki:Problem of the Day/June 24, 2011


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



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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