Difference between revisions of "2005 AIME II Problems/Problem 11"
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==Solution 3 (Telescoping)== | ==Solution 3 (Telescoping)== | ||
− | Note that <math> a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3 </math>. Then, we can generate a series of equations such that <math>\sum_{k=1}^{m-1} a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3(m-1)</math>. Then, note that all but the first and last terms on the LHS cancel out, leaving us with <math>a_m\cdot a_{m-1} - a_1\cdot a_0 = -3(m-1)</math>. Plugging in <math>a_m=0</math>, <math>a_0=37</math>, <math>a_1=72</math>, we have <math>-37\cdot 72 = -3(m-1)</math>. Solving for <math>m</math> gives <math>m=\boxed{889}</math>. | + | Note that <math> a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3 </math>. Then, we can generate a sum of series of equations such that <math>\sum_{k=1}^{m-1} a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3(m-1)</math>. Then, note that all but the first and last terms on the LHS cancel out, leaving us with <math>a_m\cdot a_{m-1} - a_1\cdot a_0 = -3(m-1)</math>. Plugging in <math>a_m=0</math>, <math>a_0=37</math>, <math>a_1=72</math>, we have <math>-37\cdot 72 = -3(m-1)</math>. Solving for <math>m</math> gives <math>m=\boxed{889}</math>. |
~sigma | ~sigma | ||
Latest revision as of 18:52, 21 November 2022
Contents
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
Solution 3 (Telescoping)
Note that . Then, we can generate a sum of series of equations such that
. Then, note that all but the first and last terms on the LHS cancel out, leaving us with
. Plugging in
,
,
, we have
. Solving for
gives
.
~sigma
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.