Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
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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 | ||
| − | < | + | <math>\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) \$<math> | + | &= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right) \$</math> |
| − | This is a [[telescope|telescoping]] series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with < | + | 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>. |
==See Also== | ==See Also== | ||
Revision as of 21:38, 10 August 2012
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 | ||
