Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
(→Solution 2) |
(→Solution 2) |
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<cmath>n=a^2+b^2c</cmath> | <cmath>n=a^2+b^2c</cmath> | ||
<cmath>\sqrt{n^2-1}=2ab\sqrt{c}</cmath> | <cmath>\sqrt{n^2-1}=2ab\sqrt{c}</cmath> | ||
− | + | calling them equations <math>A</math> and <math>B</math>, respectively. Also we have | |
− | Also we have | ||
<cmath>\frac{1}{\sqrt{n+\sqrt{n^2-1}}}=\sqrt{n-\sqrt{n^2-1}}</cmath> | <cmath>\frac{1}{\sqrt{n+\sqrt{n^2-1}}}=\sqrt{n-\sqrt{n^2-1}}</cmath> | ||
<cmath>\frac{1}{a+b\sqrt{c}}=a-b\sqrt{c}</cmath> | <cmath>\frac{1}{a+b\sqrt{c}}=a-b\sqrt{c}</cmath> | ||
− | < | + | <math></math>a^2-b^2c=1<math> |
+ | which obtains equation </math>C$. | ||
+ | L | ||
==See Also== | ==See Also== |
Revision as of 13:38, 28 November 2019
Contents
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
$$ (Error compiling LaTeX. Unknown error_msg)a^2-b^2c=1
C$.
L
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 |