Difference between revisions of "2021 Fall AMC 12B Problems/Problem 18"
(clarification) |
(Adding the See Also box) |
||
Line 32: | Line 32: | ||
-ConcaveTriangle | -ConcaveTriangle | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2021 Fall|ab=B|num-a=19|num-b=17}} | ||
+ | {{MAA Notice}} |
Revision as of 03:11, 27 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,
This means that this series approaches , as the second term is decreasing. In addition, we find 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.