AIME 2020 II · 第 6 题

Solution

Let $t_n=\frac{s_n}{5}$. Then, we have $s_n=\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$. By substitution, we find $s_3 = \frac{53}{50}$, $s_4=\frac{103}{105\cdot50}$, $s_5=\frac{101}{105}$, $s_6=100$, and $s_7=105$. So $s_n$ has a period of $5$. Thus $s_{2020}=s_5=\frac{101}{105}$. So, $\frac{101}{105\cdot 5}\implies 101+525=\boxed{626}$. ~mn28407

Solution 2 (Official MAA)

More generally, let the first two terms be $a$ and $b$ and replace $5$ and $25$ in the recursive formula by $k$ and $k^2$, respectively. Then some algebraic calculation shows that

$$t_3 = \frac{b\,k+1}{a\, k^2},~~t_4 = \frac{a\, k + b\,k+1}{a\,b\, k^3},~~ t_5 = \frac{a\,k+1}{b\, k^2},~~ t_6 = a, \text{~ and ~}t_7 =b,$$ so the sequence is periodic with period $5$. Therefore

$$t_{2020} = t_{5} = \frac{20\cdot 5 +1}{21\cdot 25} = \frac{101}{525}.$$ The requested sum is $101+525=626$.

Solution 3

Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given $t_1 = 20$ and $t_2 = 21$, so now we are able to determine the numerical value of $t_3$ using this information:

$$t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac{105 + 1}{500}t_3 = \frac{106}{500} = \frac{53}{250}$$ $$t_4 = \frac{5t_{4-1}+1}{25t_{4-2}} = \frac{5t_{3}+1}{25t_{2}} = \frac{5(\frac{53}{250}) + 1}{25(21)} = \frac{\frac{53}{50} + 1}{525} = \frac{\frac{103}{50}}{525} = \frac{103}{26250}$$ $$t_5 = \frac{5t_{5-1}+1}{25t_{5-2}} = \frac{5t_{4}+1}{25t_{3}} = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})} = \frac{\frac{103}{5250} + 1}{\frac{53}{10}} = \frac{\frac{5353}{5250}}{\frac{53}{10}} = \frac{101}{525}$$ $$t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20$$ Alas, we have figured this sequence is period 5! But since $2020 \equiv 5 \pmod 5$, we can state that $t_{2020} = t_5 = \frac{101}{525}$. According to the original problem statement, our answer is $\boxed{626}$. ~ nikenissan

Solution 4

Like solution 1, we will make a substitution. Let $u_n = 5t_n$. Then

$$u_n = \frac{25t_{n-1} + 5}{25t_{n-2}} = \frac{u_{n-1} + 1}{u_{n-2}}$$ .

Let $u_1 = a$, $u_2 = b$. The following holds for all $n \ge 3$. Then,

$$u_3 = \frac{b+1}{a}$$ $$u_4 = \frac{\frac{b+1+a}{a}}{b} = \frac{b+1+a}{ab}$$ $$u_5 = \frac{\frac{b+1+a+ab}{ab}}{\frac{b+1}{a}} = \frac{\frac{(b+1)(a+1)}{ab}}{\frac{b+1}{a}} = \frac{(b+1)(a+1)}{ab} \cdot \frac{a}{b+1} = \frac{a+1}{b}$$ $$u_6 = \frac{\frac{a+1+b}{b}}{\frac{b+1+a}{ab}} = \frac{a+1+b}{b} \cdot \frac{ab}{b+1+a} = a$$ .

Thus, $u_n$ is periodic with periodicity 5.

Note that $u_{2020} = u_5$, and thus $t_{2020} = t_5$. Also, $u_1 = 5t_1 = 100 = a$ and $u_2 = 5t_2 = 105 = b$. Then,

$$5t_5 = \frac{101}{105}$$ $$t_5 = \frac{101}{525}$$ And thus $t_{2020} = t_5 = \frac{101}{525}$. The answer is $101 + 525 = \boxed{626}$.

~Pinotation

Video Solution

https://youtu.be/_JTWJxbDC1A ~ CNCM

Video Solution 2

https://youtu.be/__B3pJMpfSk

~IceMatrix

Quick way to notice recursion loop

Round the first two values to both be 20. Then, the next element can be rounded to $\frac{1}{5}$. $t_4$ can then be quickly calculated to around $\frac{1}{250}$, and $t_5$ can be rounded to $\frac{1}{5}$. $t_6$ turns out to be around 20, which means that there is probably a loop with period 5. The rest of the solution proceeds normally.