Question

Define a sequence recursively by $x_0=5$ and

$x_{n+1}=\dfrac{x_n^2+5x_n+4}{x_n+6}$

for all nonnegative integers $n.$ Let $m$ be the least positive integer such that

$x_m\le4+\dfrac{1}{2^{20}}.$

In which of the following intervals does $m$ lie?

$[9,26]$

$[27,80]$

$[81,242]$

$[243,728]$

$[729,\infty)$

Answer: C

Concepts:recursion telescoping bounding to limit cases

Difficulty rating: 2330

Solution:

Let $a_n=x_n-4.$ A short computation gives $\begin{gathered} a_{n+1}=x_{n+1}-4 \\ =\dfrac{(x_n+5)(x_n-4)}{x_n+6} \\ =a_n\cdot\dfrac{x_n+5}{x_n+6}. \end{gathered}$

Starting from $a_0=1,$ the terms stay positive and decrease. Because $x_n$ decreases from $5$ toward $4,$ each ratio $\dfrac{x_n+5}{x_n+6}$ lies strictly between $\dfrac{9}{10}$ and $\dfrac{10}{11}.$

Hence $a_m$ is squeezed between $\left(\dfrac{9}{10}\right)^m$ and $\left(\dfrac{10}{11}\right)^m.$ Solving $a_m\le2^{-20}$ puts $m$ between about $132$ and $146,$ which lies in $[81,242].$

Thus, C is the correct answer.

2019 AMC 12B Problem 22

Answer: C

Difficulty rating: 2330

Solution:

Problem 22 in Other Years