Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
(solution) |
(tex edit... THIS WIKI IS NOT LIKE THE FORUM!) |
||
| Line 8: | Line 8: | ||
==Solution== | ==Solution== | ||
Notice that <math>\sqrt{n + \sqrt{n^2 - 1}} = \frac{1}{\sqrt{2}}\sqrt{2n + 2\sqrt{(n+1)(n-1)}} = \frac{1}{\sqrt{2}}\left(\sqrt{n+1}+\sqrt{n-1}\right)</math>. Thus, we have | Notice that <math>\sqrt{n + \sqrt{n^2 - 1}} = \frac{1}{\sqrt{2}}\sqrt{2n + 2\sqrt{(n+1)(n-1)}} = \frac{1}{\sqrt{2}}\left(\sqrt{n+1}+\sqrt{n-1}\right)</math>. Thus, we have | ||
| − | <cmath> | + | |
| − | \begin{align*} | + | <cmath>\begin{align*} |
\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}} &= \sqrt{2}\sum_{n = 1}^{9800} \frac{1}{\sqrt{n+1}+\sqrt{n-1}} \\ | \sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}} &= \sqrt{2}\sum_{n = 1}^{9800} \frac{1}{\sqrt{n+1}+\sqrt{n-1}} \\ | ||
| − | &= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right) \\ | + | &= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right) \\</cmath> |
| − | </cmath> | + | |
This is a [[telescope|telescoping]] series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with <math>\frac{1}{\sqrt{2}}\left(\sqrt{9801}+\sqrt{9800}-\sqrt{1}-\sqrt{0}\right) = 70 + 49\sqrt{2}</math>, and <math>p+q+r=\boxed{121}</math>. | This is a [[telescope|telescoping]] series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with <math>\frac{1}{\sqrt{2}}\left(\sqrt{9801}+\sqrt{9800}-\sqrt{1}-\sqrt{0}\right) = 70 + 49\sqrt{2}</math>, and <math>p+q+r=\boxed{121}</math>. | ||
Revision as of 08:09, 20 June 2008
Problem
Let 
denote the value of the sum
![\[\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}\]](http://latex.artofproblemsolving.com/0/4/2/04232c9e753b7168ec35810125ebb17786ddf7de.png)
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
\begin{align*}
\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}} &= \sqrt{2}\sum_{n = 1}^{9800} \frac{1}{\sqrt{n+1}+\sqrt{n-1}} \\
&= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right) \\ (Error compiling LaTeX. Unknown error_msg)
This is a telescoping series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with 
, and 
.
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 | ||
