2021 Fall AMC 12B Problems/Problem 18
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 1
Note that terms of the sequence lie in the interval strictly increasing.
Since the sequence tends to the limit we set
The given equation becomes from which
The given inequality becomes and we only need to consider
We have By induction, it can be proven that We substitute this into the inequality, then solve for Therefore, the least such value of is
~MRENTHUSIASM
Solution 2
If we list out the first few values of , we get the series , which always seems to be a negative power of away from . We can test this out by setting , where .
Now, we get This means that this series approaches , as the second term is decreasing. In addition, we find that .
We claim that , which can be proven by induction:
Base Case
We have .
Induction Step
Assuming that the claim is true, we have .
It follows that and . Therefore, the least value of would be .
~ConcaveTriangle
Solution 3
We are given . Multiply this equation by and subtract from both sides. The equations can then be written nicely as . Let so that .
Clearly, . Since the magnitude of is less than and because our recursive relation for squares the previous term (and negates it), we see that as . This means , so .
Isolating in our relation gives us . Substituting into the inequality, we have . Rewriting this, we get .
The sequence is much easier to handle because of its simple recursive relation. Writing out a few terms shows that . Now it just comes down to having , so .
~chezpotato
Video Solution
~MathProblemSolvingSkills.com
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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