Let ${(a_n)}_{n=0}^{\infty}$ and ${(b_n)}_{n=0}^{\infty}$ be real squences such that $a_0=40$, $b_0=41$ and for all $n\geq 0$ the given equalities hold. $$a_{n+1}=a_n+\frac{1}{b_n} \hspace{0.5 cm} \text{and} \hspace{0.5 cm} b_{n+1}=b_n+\frac{1}{a_n}$$Find the least possible positive integer value of $k$ such that the value of $a_k$ is strictly bigger than $80$.
Problem
Source: Turkey JBMO TST 2024 P6
Tags: algebra, Sequence
Assassino9931
13.05.2024 23:59
Wow, this needs a lot of experience with sequences in my opinion, much more than of a usual JBMO participant.
2460
Bravely greedily begin by changing everything in terms of $a_n$. Both sequences are positive and hence strictly increasing. We have $a_0 = 40$, $a_1 = 40\frac{1}{41}$ and $b_n = \frac{1}{a_{n+1} - a_n}$, so $\frac{1}{a_{n+2} - a_{n+1}} = \frac{1}{a_{n+1} - a_n} + \frac{1}{a_n} = \frac{a_{n+1}}{a_n(a_{n+1}-a_n)}$, thus $a_{n+2} = a_n + a_{n+1} - \frac{a_n^2}{a_{n+1}}$, equivalent to $a_{n+2}a_{n+1} - a_{n+1}a_n = a_{n+1}^2 - a_n^2$. Summing this from $0$ to $n-1$ implies $a_{n+1}a_n - a_1a_0 = a_n^2 - a_0^2$, i.e. $a_{n+1} = a_n + \frac{x}{a_n}$, where $x = a_0(a_1 - a_0) = \frac{40}{41}$.
Next, square the latter relation to obtain $a_{n+1}^2 = a_n^2 + 2x + \frac{x^2}{a_n^2}$ and hence $a_n^2 = a_0^2 + 2nx + x^2\sum_{i=0}^{n-1}\frac{1}{a_i^2}$. In particular, if $n$ is such that $a_0^2 + 2nx \geq 80^2$, then $a_n > 80$ would hold; now $\frac{80}{41} \cdot n \geq 80^2 - 40^2 = 60 \cdot 80$ holds precisely for $n\geq 2460$. On the other hand, if $n \leq 2459$, then (as the sequence is strictly increasing) $a_n^2 \leq a_0^2 + 2nx + x^2 \cdot \frac{n}{a_0^2} = 40^2 + 2 \cdot 2459 \cdot \frac{40}{41} + 2459(a_1-a_0)^2$ which is less than $80^2$, since $2 \cdot 2459 \cdot \frac{40}{41} + \frac{2459}{41^2} = \frac{2459 \cdot 3281}{41^2} < 4800$ by direct calculation (no idea if it can be shortened).
Could this be done by working with $a_n < b_n$ and $a_{n+1}b_{n+1} = a_nb_n + 2 + \frac{1}{a_nb_n}$? Tried a bit and did not succeed, too lazy to try again.
D.K.K
14.05.2024 21:00
$\frac{a_{n+1}}{b_{n+1}} = \frac{a_{n} + \frac{1}{b_{n}}}{b_{n} + \frac{1}{a_{n}}} = \frac{a_{n} + \frac{1}{b_{n}}}{b_{n} + \frac{1}{a_{n}}} \cdot \frac{b_n}{a_n} \cdot \frac{a_n}{b_n} = \frac{a_n}{b_n}$ This shows that $\frac{a_n}{b_n}$ is constant. Therefore if $a_k > 80 = 40 \cdot 2$ then $b_k > 82 = 41 \cdot 2$ Multiplying, $a_k \cdot b_k = a_{k - 1} \cdot b_{k - 1} + \frac{1}{a_{k - 1} \cdot b_{k - 1}} + 2 > 80 \cdot 82 = 6560$ we can ignore $ \frac{1}{a_{k - 1} \cdot b_{k - 1}}$ since it is smaller than 2 when combined. we know that $a_0 \cdot b_0 = 40 \cdot 41 = 1640$ Thus, $k = \frac{6560 - 1640}{2} = 2460$ I couldn't see this solution in the exam. I stated that $b_n > a_n + 1$ and found $2420$.
Z4ADies
02.06.2024 11:52
head solved in 3 mins. First two statements roughly equal to $a_{n+1}b_{n}=a_{n}b_{n}+1=b_{n+1}a_{n}=b_{n}a_{n}+1$ so,$\frac{b_{n}}{a_{n}}=\frac{b_{n+1}}{a_{n+1}}$ Let's use this equality in first equation . We know that,$\frac{b_{n}}{a_{n}}=\frac{40}{41}$.Now it transforms like $a_{n+1}=a_{n}+\frac{40}{41a_{n}}$ which is equivalent to $a^2_{n+1}=a^2_{n}+(\frac{40}{41a_{n}})^2+\frac{80}{41}$. Telescopic summation gives us. $a^2_{n+1}>a^2_{0}+\frac{(n+1)80}{41}>6400$ so, we can easily found $n=2460$
wizixez
31.12.2024 11:16
Bruw easiest i did not expected this to be this easy Main claim is that $\forall n\ge 0 \frac{a_n}{b_n}=\frac{40}{41}$ rest is just inequalities etc.