Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 Fall AMC 12B Problems/Problem 18"
m (→‎Solution)
Line 7: Line 7:
 
<math>\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401</math>
 
<math>\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401</math>
  
−
==Solution==
+
==Solution 1==
  
 
If we list out the first few values of <math>k</math>, we get the series <math>\frac{1}{4}, \frac{3}{8}, \frac{15}{32}, \frac{255}{512}</math>, which seem to always be a negative power of <math>2</math> away from <math>\frac{1}{2}</math>. We can test this out by setting <math>u_k</math> to <math>\frac{1}{2}-\frac{1}{2^{n_k}}</math>.  
 
If we list out the first few values of <math>k</math>, we get the series <math>\frac{1}{4}, \frac{3}{8}, \frac{15}{32}, \frac{255}{512}</math>, which seem to always be a negative power of <math>2</math> away from <math>\frac{1}{2}</math>. We can test this out by setting <math>u_k</math> to <math>\frac{1}{2}-\frac{1}{2^{n_k}}</math>.  

Revision as of 17:56, 17 March 2022

Problem

Set $u_0 = \frac{1}{4}$, and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence \[u_{k+1} = 2u_k - 2u_k^2.\]

This sequence tends to a limit; call it $L$. What is the least value of $k$ such that \[|u_k-L| \le \frac{1}{2^{1000}}?\]

$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(D)}\: 329\qquad\textbf{(E)}\: 401$

Solution 1

If we list out the first few values of $k$, we get the series $\frac{1}{4}, \frac{3}{8}, \frac{15}{32}, \frac{255}{512}$, which seem to always be a negative power of $2$ away from $\frac{1}{2}$. We can test this out by setting $u_k$ to $\frac{1}{2}-\frac{1}{2^{n_k}}$.

Now, \begin{align*} u_{k+1} &= 2\cdot\left(\frac{1}{2}-\frac{1}{2^{n_{k}}}\right)-2\cdot\left(\frac{1}{2}-\frac{1}{2^{n_{k}}}\right)^2 \\ &= 1-\frac{1}{2^{n_k - 1}}-2\cdot\left(\frac{1}{4}-\frac{1}{2^{n_k}}+\frac{1}{2^{2 \cdot n_k}}\right)\\ &= 1-\frac{1}{2^{n_k - 1}}-\frac{1}{2}+\frac{1}{2^{n_k-1}}-\frac{1}{2^{2 \cdot n_k-1}} \\ &= \frac{1}{2}-\frac{1}{2^{2 \cdot n_k-1}} \\ \end{align*}

This means that this series approaches $\frac{1}{2}$, as the second term is decreasing. In addition, we find that $n_{k+1}=2 \cdot n_k-1$.

We see that $n_k$ seems to always be $1$ above a power of $2$. We can prove this using induction.

Claim

$n_k = 2^k+1$

Base case

We have $n_0=2^0+1$, which is true.

Induction Step

Assuming that the claim is true, we have $n_{k+1}=2 \cdot 2^k+2-1=2^{k+1}+1$.

It follows that $n_{10}=2^{10}+1>1000$, and $n_9=2^9+1<1000$. Therefore, the least value of $k$ would be $\boxed{\textbf{(A) }10}$.

-ConcaveTriangle

See Also

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_18&oldid=172719"