AIME 2005 II · 第 11 题

Solution 1

For $0 < k < m$, we have

$a_{k}a_{k+1} = a_{k-1}a_{k} - 3$.

Thus the product $a_{k}a_{k+1}$ is a monovariant: it decreases by 3 each time $k$ increases by 1. For $k = 0$ we have $a_{k}a_{k+1} = 37\cdot 72$, so when $k = \frac{37 \cdot 72}{3} = 888$, $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = \boxed{889}$, our answer.

Note: In order for $a_{m} = 0$ we need $a_{m-2}a_{m-1}=3$ simply by the recursion definition.

Solution 2

Plugging in $k = m-1$ to the given relation, we get $0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}$. Inspecting the value of $a_{k}a_{k+1}$ for small values of $k$, we see that $a_{k}a_{k+1} = 37\cdot 72 - 3k$. Setting the RHS of this equation equal to $3$, we find that $m$ must be $\boxed{889}$.

~ anellipticcurveoverq

Induction Proof

As above, we experiment with some values of $a_{k}a_{k+1}$, conjecturing that $a_{m-p}a_{m-p-1}$ = $3p$ ,where $m$ is a positive integer and so is $p$, and we prove this formally using induction. The base case is for $p = 1$, $a_{m} = a_{m-2} - 3/a_{m-1}$ Since $a_{m} = 0$, $a_{m-1}a_{m-2} = 3$; from the recursion given in the problem $a_{m-p+1} = a_{m-p-1} - 3/a_{m-p}$, so $a_{m-p+1} = 3p/a_{m-p} - 3/a_{m-p} = 3(p-1)/a_{m-p}$, so $a_{m-p}a_{m-p+1} = a_{m-(p-1)}a_{m-(p-1)-1} = 3(p-1)$, hence proving our formula by induction. ~USAMO2023

Solution 3 (Telescoping)

Note that $a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3$. Then, we can generate a sum of series of equations such that $\sum_{k=1}^{m-1} a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3(m-1)$. Then, note that all but the first and last terms on the LHS cancel out, leaving us with $a_m\cdot a_{m-1} - a_1\cdot a_0 = -3(m-1)$. Plugging in $a_m=0$, $a_0=37$, $a_1=72$, we have $-37\cdot 72 = -3(m-1)$. Solving for $m$ gives $m=\boxed{889}$. ~sigma

Video solution

https://www.youtube.com/watch?v=JfxNr7lv7iQ