Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:
Let $a_{1}, a_{2}, \ldots$ be a sequence defined by $a_{1}=a_{2}=1$ and $a_{n+2}=a_{n+1}+a_{n}$ for $n \geq 1$ . Find
$\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}$

Solution

Solution:
Let $X$ denote the desired sum. Note that
$\begin{array}{rl} X & = \frac{1}{4^{2}}+\frac{1}{4^{3}}+\frac{2}{4^{4}}+\frac{3}{4^{5}}+\frac{5}{4^{6}}+\ldots \\ 4 X & =\quad \frac{1}{4^{1}}+\frac{1}{4^{2}}+\frac{2}{4^{3}}+\frac{3}{4^{4}}+\frac{5}{4^{5}}+\frac{8}{4^{6}}+\ldots \\ 16 X &=\frac{1}{4^{0}}+\frac{1}{4^{1}}+\frac{2}{4^{2}}+\frac{3}{4^{3}}+\frac{5}{4^{4}}+\frac{8}{4^{5}}+\frac{13}{4^{6}}+\ldots \end{array}$
so that $X+4 X=16 X-1$ , and $X=1 / 11$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.