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AoPS Wiki talk:Problem of the Day/September 10, 2011

Solution

Let \[S=\frac{F_1}{3}+\frac{F_2}{9}+\frac{F_3}{27}+\frac{F_4}{81}+\frac{F_5}{243}+\cdots\] Multiplying through by 1/3, we see \[\frac{1}{3}S=\frac{F_1}{9}+\frac{F_2}{27}+\frac{F_3}{81}+\frac{F_4}{243}+\cdots\] and, noting that $F_{n+1}-F_n=F_{n-1}$, we subtract: \[\frac{2}{3}S=\frac{1}{3}+\frac{F_1}{27}+\frac{F_2}{81}+\frac{F_3}{243}+\cdots\] and, multiplying through by 9, we see S in the right side: \[6S=3+\frac{F_1}{3}+\frac{F_2}{9}+\frac{F_3}{27}+\frac{F_4}{81}+\frac{F_5}{243}+\cdots=3+S\] and thus $5S=3$ and $\boxed{S=\frac{3}{5}}$.

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