Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
(solution)
Line 2: Line 2:
 
Let <math>S</math> denote the value of the sum
 
Let <math>S</math> denote the value of the sum
  
−
<math>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</math>
+
<cmath>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</cmath>
  
 
<math>S</math> can be expressed as <math>p + q \sqrt{r}</math>, where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime. Determine <math>p + q + r</math>.
 
<math>S</math> can be expressed as <math>p + q \sqrt{r}</math>, where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime. Determine <math>p + q + r</math>.
  
 
==Solution==
 
==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
 +
<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}} \\
 +
&= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right) \\
 +
</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>.
  
 
==See also==
 
==See also==
 +
{{Mock AIME box|year=Pre 2005|n=3|num-b=5|num-a=7|source=19349}}
 +
 +
[[Category:Intermediate Algebra Problems]]

Revision as of 21:12, 19 June 2008

Problem

Let $S$ denote the value of the sum

\[\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}\]

$S$ can be expressed as $p + q \sqrt{r}$, where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime. Determine $p + q + r$.

Solution

Notice that $\sqrt{n + \sqrt{n^2 - 1}} = \frac{1}{\sqrt{2}}\sqrt{2n + 2\sqrt{(n+1)(n-1)}} = \frac{1}{\sqrt{2}}\left(\sqrt{n+1}+\sqrt{n-1}\right)$. 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 $\frac{1}{\sqrt{2}}\left(\sqrt{9801}+\sqrt{9800}-\sqrt{1}-\sqrt{0}\right) = 70 + 49\sqrt{2}$, and $p+q+r=\boxed{121}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_3_Pre_2005_Problems/Problem_6&oldid=26701"