# 1964 AHSME Problems/Problem 31

## Problem

Let $$f(n)=\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n.$$

Then $f(n+1)-f(n-1)$, expressed in terms of $f(n)$, equals: $\textbf{(A) }\frac{1}{2}f(n)\qquad\textbf{(B) }f(n)\qquad\textbf{(C) }2f(n)+1\qquad\textbf{(D) }f^2(n)\qquad \textbf{(E) }$ $\frac{1}{2}(f^2(n)-1)$

## Solution

We compute $f(n+1)$ and $f(n-1)$, while pulling one copy of the exponential part outside: $f(n+1) = \dfrac{5+3\sqrt{5}}{10}\cdot \frac{1 + \sqrt{5}}{2}\cdot\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\cdot \frac{1 - \sqrt{5}}{2}\cdot\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ $f(n+1) = \dfrac{20+8\sqrt{5}}{20}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{20-8\sqrt{5}}{20}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ $f(n-1) = \dfrac{5+3\sqrt{5}}{10}\cdot \frac{2}{1 + \sqrt{5}}\cdot\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\cdot \frac{2}{1 - \sqrt{5}}\cdot\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ $f(n-1) = \dfrac{(5+3\sqrt{5})(1 - \sqrt{5})}{5(1 + \sqrt{5})(1 - \sqrt{5})}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{(5-3\sqrt{5})(1 + \sqrt{5})}{5(1 - \sqrt{5})(1 + \sqrt{5})}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ $f(n-1) = \dfrac{-10-2\sqrt{5}}{5(-4)}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{-10 + 2\sqrt{5}}{5(-4)}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ $f(n-1) = \dfrac{10+2\sqrt{5}}{20}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{10 - 2\sqrt{5}}{20}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$

Computing $f(n+1) - f(n-1)$ gives: $f(n+1) - f(n-1) = \dfrac{10+6\sqrt{5}}{20}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{10-6\sqrt{5}}{20}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ $f(n+1) - f(n-1) = \dfrac{5+3\sqrt{5}}{10}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-3\sqrt{5}}{10}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ $f(n+1) - f(n-1) = f(n)$

Thus, the answer is $\boxed{\textbf{(B)}}$.

## Solution 2

Notice that $\left(\frac{1+\sqrt{5}}{2}\right)^n$ and $\left(\frac{1-\sqrt{5}}{2}\right)^n$ are the characteristics roots for the recurrence relation $F_n = F_{n-1} + F_{n-2}$ (think about Binet's formula). And $f(n)$ is the solution (i.e. $a_n$) to the recurrence relation with constants $a = \frac{5+3\sqrt{5}}{10}$ and $b = \frac{5-3\sqrt{5}}{10}$. Thus, $f(n+1) - f(n-1) = f(n)$, and the answer is $\boxed{\textbf{(B)}}$. -nullptr07

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 