Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
(→Solution 2) |
(→Solution 2) |
||
Line 31: | Line 31: | ||
<cmath>\sqrt{n+\sqrt{n^2-1}}=a+b\sqrt{c}</cmath> | <cmath>\sqrt{n+\sqrt{n^2-1}}=a+b\sqrt{c}</cmath> | ||
<cmath>\sqrt{n-\sqrt{n^2-1}}=a-b\sqrt{c}</cmath> | <cmath>\sqrt{n-\sqrt{n^2-1}}=a-b\sqrt{c}</cmath> | ||
− | for some <math>a</math>, <math>b</math>, <math>c</math>. | + | for some <math>a</math>, <math>b</math>, <math>c</math>. Then we have |
+ | <cmath>\frac{1}{a+b\sqrt{c}}=a-b\sqrt{c}</cmath> | ||
+ | <cmath>a^2-b^2\cdot c=1</cmath> | ||
==See Also== | ==See Also== |
Revision as of 12:18, 28 November 2019
Contents
[hide]Problem
Let denote the value of the sum
can be expressed as , where and are positive integers and is not divisible by the square of any prime. Determine .
Solution
Notice that . Thus, we have
This is a telescoping series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with , and .
Solution 2
Simplifying the expression yields Now we can assume that for some , , . Then we have
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |