Difference between revisions of "2005 AIME II Problems/Problem 11"
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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. | 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 | ~USAMO2023 | ||
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+ | ==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 -37\cdot 72 = -3(m-1)<math>. Solving for </math>m<math> gives </math>m=\boxed{889}$. | ||
+ | ~sigma | ||
+ | |||
==Video solution== | ==Video solution== |
Revision as of 17:51, 21 November 2022
Contents
[hide]Problem
Let be a positive integer, and let be a sequence of reals such that and for Find
Solution 1
For , we have
.
Thus the product is a monovariant: it decreases by 3 each time increases by 1. For we have , so when , will be zero for the first time, which implies that , our answer.
Note: In order for we need simply by the recursion definition.
Solution 2
Plugging in to the given relation, we get . Inspecting the value of for small values of , we see that . Setting the RHS of this equation equal to , we find that must be .
~ anellipticcurveoverq
Induction Proof
As above, we experiment with some values of , conjecturing that = ,where is a positive integer and so is , and we prove this formally using induction. The base case is for , Since , ; from the recursion given in the problem , so , so , hence proving our formula by induction. ~USAMO2023
Solution 3 (Telescoping)
Note that . Then, we can generate a series of equations such that . Then, note that all but the first and last terms on the LHS cancel out, leaving us with . Plugging in , , , we have -37\cdot 72 = -3(m-1)mm=\boxed{889}$. ~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 |
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