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2019 AMC 10A Problems/Problem 15: Difference between revisions

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<math>\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}</math>
<math>\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}</math>


<math>b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}</math>
<math>b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}</math>=4b_{n-3}-3b_{n-4}=\dots<math>


By recursively following this pattern, we can see that <math>b_n=(n-1) \cdot b_2 - (n-1) \cdot b_1</math>.
By recursively following this pattern, we can see that </math>b_n=(n-1) \cdot b_2 - (n-1) \cdot b_1<math>.


By plugging in 2019, we get: <math>b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8705}{3}</math>. Since they are relatively prime, the answer is <math>8708</math> which implies <math>\rightarrow \boxed{E}</math>
By plugging in 2019, we get: </math>b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8705}{3}<math>. Since they are relatively prime, the answer is </math>8708<math> which implies </math>\rightarrow \boxed{E}$


==See Also==
==See Also==

Revision as of 17:40, 9 February 2019

The following problem is from both the 2019 AMC 10A #15 and 2019 AMC 12A #9, so both problems redirect to this page.

Problem

A sequence of numbers is defined recursively by $a_1 = 1$, $a_2 = \frac{3}{7}$, and \[a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}\]for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive inegers. What is $p+q ?$

$\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078$

Solution 1

Using the recursive formula, we find $a_3=\frac{3}{11}$, $a_4=\frac{3}{15}$, and so on. It appears that $a_n=\frac{3}{4n-1}$, for all $n$. Setting $n=2019$, we find $a_{2019}=\frac{3}{8075}$, so the answer is $\boxed{\textbf{(E) }8078}$.


To prove this formula, we use induction. We are given that $a_1=1$ and $a_2=\frac{3}{7}$, which satisfy our formula. Now assume the formula holds true for all $n\le m$ for some positive integer $m$. By our assumption, $a_{m-1}=\frac{3}{4m-5}$ and $a_m=\frac{3}{4m-1}$. Using the recursive formula, \[a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}=\frac{(\frac{3}{4m-5}\cdot\frac{3}{4m-1})(4m-5)(4m-1)}{(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1})(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},\] so our induction is complete.

Solution 2

Since we are finding the sum of the numerator and the denominator, consider the function $b_n = \frac{1}{a_n}$.

$\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}$

$b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}$=4b_{n-3}-3b_{n-4}=\dots$By recursively following this pattern, we can see that$b_n=(n-1) \cdot b_2 - (n-1) \cdot b_1$.

By plugging in 2019, we get:$ (Error compiling LaTeX. Unknown error_msg)b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8705}{3}$. Since they are relatively prime, the answer is$8708$which implies$\rightarrow \boxed{E}$

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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