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

m (→‎Solution)
Line 11: Line 11:
 
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>.  
  
−
Now,  
+
Now, we get
 
<cmath>\begin{align*}
 
<cmath>\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 \\
 
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}}-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}} \\
 
&= 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}} \\
+
&= \frac{1}{2}-\frac{1}{2^{2 \cdot n_k-1}}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
 
 
This means that this series approaches <math>\frac{1}{2}</math>, as the second term is decreasing. In addition, we find that <math>n_{k+1}=2 \cdot n_k-1</math>.  
 
This means that this series approaches <math>\frac{1}{2}</math>, as the second term is decreasing. In addition, we find that <math>n_{k+1}=2 \cdot n_k-1</math>.  
  
Line 35: Line 34:
 
Assuming that the claim is true, we have <math>n_{k+1}=2 \cdot 2^k+2-1=2^{k+1}+1</math>.
 
Assuming that the claim is true, we have <math>n_{k+1}=2 \cdot 2^k+2-1=2^{k+1}+1</math>.
  
−
It follows that <math>n_{10}=2^{10}+1>1000</math>, and <math>n_9=2^9+1<1000</math>. Therefore, the least value of <math>k</math> would be <math>\boxed{\textbf{(A) }10}</math>.
+
It follows that <math>n_{10}=2^{10}+1>1000</math>, and <math>n_9=2^9+1<1000</math>. Therefore, the least value of <math>k</math> would be <math>\boxed{\textbf{(A)}\: 10}</math>.
 +
 
 +
~ConcaveTriangle
 +
 
 +
==Solution 2==
 +
 
 +
Note that all terms of the sequence <math>(u_k)</math> lie in the interval <math>\left(0,\frac12\right),</math> strictly increasing.
  
−
-ConcaveTriangle
+
Since the sequence <math>(u_k)</math> tends to the limit <math>L,</math> we set <math>u_{k+1}=u_k=L>0.</math> The given equation becomes <cmath>L=2L-2L^2,</cmath> from which <math>L=\frac12.</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=19|num-b=17}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=19|num-b=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:36, 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, 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 all 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 $L=\frac12.$

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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