2005 AIME II Problems/Problem 11
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 | ||
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