Question

Define a sequence recursively by $x_0=5$ and $x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}$ for all nonnegative integers $n.$ Let $m$ be the least positive integer such that $x_m\leq 4+\frac{1}{2^{20}}.$ In which of the following intervals does $m$ lie?

$[9,26]$

$[27,80]$

$[81,242]$

$[243,728]$

$[729,\infty)$

Answer: C

Concepts:recursion inequality bounding to limit cases

Difficulty rating: 2380

Solution:

Let $a_n=x_n-4$ . Then $a_0=1$ , and simplifying the recurrence gives $a_{n+1}=\frac{a_n(a_n+9)}{a_n+10}.$ As long as $0<a_n\le1$ , this implies $\frac9{10}a_n\le a_{n+1}\le\frac{10}{11}a_n.$

By induction, $\left(\dfrac9{10}\right)^n\le a_n\le\left(\dfrac{10}{11}\right)^n$ . For $n=80$ , $\left(\dfrac{10}{9}\right)^{80}<2^{20}$ , so $\left(\dfrac9{10}\right)^{80}>2^{-20}$ , and therefore $m>80$ .

Also $\left(\dfrac{11}{10}\right)^8>2$ , so $\left(\dfrac{10}{11}\right)^{160}<2^{-20}$ , which gives $m\le160$ . Hence $81\le m\le160$ , so $m$ lies in $[81,242]$ . Thus, C is the correct answer.

2019 AMC 10B Problem 24

Answer: C

Difficulty rating: 2380

Solution:

Problem 24 in Other Years