Difference between revisions of "2002 AIME I Problems/Problem 4"
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== Solution 1 == | == Solution 1 == | ||
− | <math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus, | + | Using [[partial fraction decomposition]] yields <math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus, |
<math>a_m+a_{m+1}+\cdots +a_{n-1}=\dfrac{1}{m}-\dfrac{1}{m+1}+\dfrac{1}{m+1}-\dfrac{1}{m+2}+\cdots +\dfrac{1}{n-1}-\dfrac{1}{n}=\dfrac{1}{m}-\dfrac{1}{n}</math> | <math>a_m+a_{m+1}+\cdots +a_{n-1}=\dfrac{1}{m}-\dfrac{1}{m+1}+\dfrac{1}{m+1}-\dfrac{1}{m+2}+\cdots +\dfrac{1}{n-1}-\dfrac{1}{n}=\dfrac{1}{m}-\dfrac{1}{n}</math> | ||
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*If <math>m=29t</math>, a similar argument to the one above implies <math>m=29(28)</math> and <math>n=28</math>, which implies <math>m>n</math>. This is impossible since <math>n-m>0</math>. | *If <math>m=29t</math>, a similar argument to the one above implies <math>m=29(28)</math> and <math>n=28</math>, which implies <math>m>n</math>. This is impossible since <math>n-m>0</math>. | ||
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== Solution 2 == | == Solution 2 == |
Latest revision as of 10:27, 5 October 2024
Problem
Consider the sequence defined by for . Given that , for positive integers and with , find .
Solution 1
Using partial fraction decomposition yields . Thus,
Which means that
Since we need a factor of 29 in the denominator, we let .* Substituting, we get
so
Since is an integer, , so . It quickly follows that and , so .
*If , a similar argument to the one above implies and , which implies . This is impossible since .
Solution 2
Note that . This can be proven by induction. Thus, . Cross-multiplying yields , and adding to both sides gives . Clearly, and . Hence, , , and .
~ keeper1098
Video Solution by OmegaLearn
https://youtu.be/lH-0ul1hwKw?t=134
~ pi_is_3.14
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.