Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
(→Solution 2) |
(→Solution 2) |
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<cmath>2a^2=n+1</cmath> | <cmath>2a^2=n+1</cmath> | ||
<cmath>a=\sqrt{\frac{n+1}{2}}</cmath> | <cmath>a=\sqrt{\frac{n+1}{2}}</cmath> | ||
+ | Squaring equation <math>B</math> and substituting yields | ||
+ | <cmath>4a^2b^2c=n^2-1</cmath> | ||
+ | <cmath>2\cdot(n+1)\cdot b^2c=n^2-1</cmath> | ||
+ | <cmath>b^2c=\frac{(n-1)(n+1)}{2\cdot(n+1)}</cmath> | ||
==See Also== | ==See Also== |
Revision as of 14:16, 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 , , .
Squaring the first equation yields which gives the system of equations calling them equations and , respectively.
Also we have which obtains equation .
Adding equations and yields Squaring equation and substituting yields
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 |