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

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

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2021 Fall AMC 12B Problems/Problem 18

Problem

Solution 1

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