Difference between revisions of "2021 Fall AMC 12B Problems/Problem 18"
m (→Solution) |
m (→Solution) |
||
Line 13: | Line 13: | ||
Now, | Now, | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | u_{k+1} &= 2\cdot(\frac{1}{2}-\frac{1}{2^{n_k} | + | 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(\frac{1}{4}-\frac{1}{2^{n_k}}+\frac{1}{2^{2 \cdot n_k}}) \\ | + | &= 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}} \\ |
Revision as of 18:36, 24 November 2021
Problem
Set , and for
let
be determined by the recurrence
This sequence tends to a limit; call it . What is the least value of
such that
Solution
If we list out the first few values of k, we get the series , which seem to always be a negative power of 2 away from
. We can test this out by setting
to
.
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*} (Error compiling LaTeX. Unknown error_msg)
This means that .
We see that seems to always be
above a power of
. We can prove this using induction.
Claim:
Base case:
Induction:
It follows that , and
. Therefore, the least value of
would be
.
-ConcaveTriangle