Difference between revisions of "2005 AIME II Problems/Problem 11"
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+ | ==Induction Proof== | ||
+ | As above, we experiment with some values of <math>a_{k}a_{k+1}</math>, conjecturing that <math>a_{m-p}a_{m-p-1}</math> = <math>3p</math> ,where <math>m</math> is a positive integer and so is <math>p</math>, and we prove this formally using induction. The base case is for <math>p = 1</math>, <math>a_{m} = a_{m-2} - 3/a_{m-1}</math> Since <math>a_{m} = 0 </math>, <math>a_{m-1}a_{m-2} = 3</math>; from the recursion given in the problem <math>a_{m-p+1} = a_{m-p-1} - 3/a_{m-p}</math>, so <math>a_{m-p+1} = 3p/a_{m-p} - 3/a_{m-p} = 3(p-1)/a_{m-p}</math>, so <math>a_{m-p}a_{m-p+1} = a_{m-(p-1)}a_{m-(p-1)-1} = 3(p-1)</math>, hence proving our formula by induction. | ||
+ | ~USAMO2023 | ||
==Video solution== | ==Video solution== |
Revision as of 10:45, 7 July 2022
Problem
Let be a positive integer, and let
be a sequence of reals such that
and
for
Find
Solution 1
For , we have
.
Thus the product is a monovariant: it decreases by 3 each time
increases by 1. For
we have
, so when
,
will be zero for the first time, which implies that
, our answer.
Note: In order for we need
simply by the recursion definition.
Solution 2
Plugging in to the given relation, we get
. Inspecting the value of
for small values of
, we see that
. Setting the RHS of this equation equal to
, we find that
must be
.
~ anellipticcurveoverq
Induction Proof
As above, we experiment with some values of , conjecturing that
=
,where
is a positive integer and so is
, and we prove this formally using induction. The base case is for
,
Since
,
; from the recursion given in the problem
, so
, so
, hence proving our formula by induction.
~USAMO2023
Video solution
https://www.youtube.com/watch?v=JfxNr7lv7iQ
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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