2021 Fall AMC 12B Problems/Problem 18

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

Solution 2

Note that terms of the sequence $(u_k)$ lie in the interval $\left(0,\frac12\right),$ strictly increasing.

Since the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$

The given equation becomes $L=2L-2L^2,$ from which $L=\frac12.$

The given inequality becomes $\frac12-\frac{1}{2^{1000}} \leq u_k \leq \frac12+\frac{1}{2^{1000}},$ and we only need to consider $\frac12-\frac{1}{2^{1000}} \leq u_k.$

We have $\begin{alignat*}{8} u_0 &= \frac14 &&= \frac{2^1-1}{2^2}, \\ u_1 &= \frac38 &&= \frac{2^2-1}{2^3}, \\ u_2 &= \frac{15}{32} &&= \frac{2^4-1}{2^5}, \\ u_3 &= \frac{255}{512} &&= \frac{2^8-1}{2^9}, \\ & \ \vdots \end{alignat*}$

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

Page

Toolbox

Search

2021 Fall AMC 12B Problems/Problem 18

Contents

Problem

Solution 1

Solution 2

See Also