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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
  
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<math>\begin{align*}
+
<cmath>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</cmath>
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\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}} \\
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<cmath>= \sqrt{2}\sum_{n = 1}^{9800} \frac{1}{\sqrt{n+1}+\sqrt{n-1}}</cmath>
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&= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right) \&#036;</math>
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<cmath>= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right)</cmath>
  
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This is a [[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>.
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This is a [[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 01:56, 19 March 2015

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

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

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)
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Problem 5 		Followed by
Problem 7
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