Competition problem — Algebra — Maths Olympiad Prep

Library / /346 of 860

Algebra Difficulty 5.1 AIME, harder Find the answer

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 $\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}$

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

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

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.