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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 18"

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Revision as of 02:11, 27 November 2021

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

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: $n_0=2^0+1$

Induction: $n_{k+1}=2 \cdot 2^k+2-1=2^{k+1}+1$

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

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