Difference between revisions of "2019 AMC 10A Problems/Problem 15"
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Using the recursive formula, we find <math>a_3=\frac{3}{11}</math>, <math>a_4=\frac{3}{15}</math>, and so on. It appears that <math>a_n=\frac{3}{4n-1}</math>, for all <math>n</math>. Setting <math>n=2019</math>, we find <math>a_{2019}=\frac{3}{8075}</math>, so the answer is <math>\boxed{\textbf{(E) }8078}</math>. | Using the recursive formula, we find <math>a_3=\frac{3}{11}</math>, <math>a_4=\frac{3}{15}</math>, and so on. It appears that <math>a_n=\frac{3}{4n-1}</math>, for all <math>n</math>. Setting <math>n=2019</math>, we find <math>a_{2019}=\frac{3}{8075}</math>, so the answer is <math>\boxed{\textbf{(E) }8078}</math>. | ||
− | + | To prove this formula, we use induction. We are given that <math>a_1=1</math> and <math>a_2=\frac{3}{7}</math>, which satisfy our formula. Now assume the formula holds true for all <math>n\le m</math> for some positive integer <math>m</math>. By our assumption, <math>a_{m-1}=\frac{3}{4m-5}</math> and <math>a_m=\frac{3}{4m-1}</math>. Using the recursive formula, <cmath>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{\left(\frac{3}{4m-5}\cdot\frac{3}{4m-1}\right)(4m-5)(4m-1)}{\left(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}\right)(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},</cmath> | |
− | To prove this formula, we use induction. We are given that <math>a_1=1</math> and <math>a_2=\frac{3}{7}</math>, which satisfy our formula. Now assume the formula holds true for all <math>n\le m</math> for some positive integer <math>m</math>. By our assumption, <math>a_{m-1}=\frac{3}{4m-5}</math> and <math>a_m=\frac{3}{4m-1}</math>. Using the recursive formula, <cmath>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},</cmath> | ||
so our induction is complete. | so our induction is complete. | ||
==Solution 2== | ==Solution 2== | ||
− | Since we are finding the sum of the numerator and the denominator, consider the | + | Since we are interested in finding the sum of the numerator and the denominator, consider the sequence defined by <math>b_n = \frac{1}{a_n}</math>. |
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− | <math>b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}=4b_{n-3}-3b_{n-4}=\dots</math> | + | We have <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>, so in other words, <math>b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}=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-2) \cdot b_1</math>. | By recursively following this pattern, we can see that <math>b_n=(n-1) \cdot b_2 - (n-2) \cdot b_1</math>. | ||
− | By plugging in 2019, we | + | By plugging in 2019, we thus find <math>b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8075}{3}</math>. Since the numerator and the denominator are relatively prime, the answer is <math>\boxed{\textbf{(E) } 8078}</math>. |
==See Also== | ==See Also== |
Revision as of 00:57, 27 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.
Contents
Problem
A sequence of numbers is defined recursively by , , and for all Then can be written as , where and are relatively prime positive integers. What is
Solution 1
Using the recursive formula, we find , , and so on. It appears that , for all . Setting , we find , so the answer is .
To prove this formula, we use induction. We are given that and , which satisfy our formula. Now assume the formula holds true for all for some positive integer . By our assumption, and . Using the recursive formula, so our induction is complete.
Solution 2
Since we are interested in finding the sum of the numerator and the denominator, consider the sequence defined by .
We have , so in other words, .
By recursively following this pattern, we can see that .
By plugging in 2019, we thus find . Since the numerator and the denominator are relatively prime, the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.