Mock AIME 3 Pre 2005 Problems/Problem 6
Contents
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
![\[= \sqrt{2}\sum_{n = 1}^{9800} \frac{1}{\sqrt{n+1}+\sqrt{n-1}}\]](http://latex.artofproblemsolving.com/f/4/9/f496ffe5cd7a9cc8dbe2f46f04cce9fe131301cb.png)
![\[= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right)\]](http://latex.artofproblemsolving.com/b/4/c/b4c362df83e615c24ee647f57ba1180a897ea49c.png)
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
\[\begin{\align*}
S &= \sum_{n=1}^{9800}\frac{1}{\sqrt{n+\sqrt{n^2-1}}} \\
&= \sum_{n=1}^{9800}\frac{\sqrt{n+\sqrt{n^2-1}}}{n+\sqrt{n^2-1}} \\
&= \sum_{n=1}^{9800}(\frac{\sqrt{n+\sqrt{n^2-1}}}{n+\sqrt{n^2-1}})\cdot(\frac{n-\sqrt{n^2-1}}{n-\sqrt{n^2-1}}) \\
&= \sum_{n=1}^{9800}\frac{(\sqrt{n+\sqrt{n^2-1})\cdot(\sqrt{n+\sqrt{n^2-1})^2}{\sqrt{n+\sqrt{n^2-1}}} \\
\end{\align*}\] (Error compiling LaTeX. Unknown error_msg)
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 | ||
