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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"

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&= \sum_{n=1}^{9800}\sqrt{n-\sqrt{n^2-1}}
 
&= \sum_{n=1}^{9800}\sqrt{n-\sqrt{n^2-1}}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
Now we can assume that
 +
<cmath>\sqrt{n+\sqrt{n^2-1}}=a+b\sqrt{c}</cmath>
 +
<cmath>\sqrt{n-\sqrt{n^2-1}}=a-b\sqrt{c}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 12:13, 28 November 2019

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 $p+q+r=\boxed{121}$.


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}(\sqrt{n+\sqrt{n^2-1}})\cdot(\sqrt{n+\sqrt{n^2-1}})^2 \\ &= \sum_{n=1}^{9800}\sqrt{n-\sqrt{n^2-1}} \end{align*} Now we can assume that \[\sqrt{n+\sqrt{n^2-1}}=a+b\sqrt{c}\] \[\sqrt{n-\sqrt{n^2-1}}=a-b\sqrt{c}\]

See Also

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by
Problem 5 		Followed by
Problem 7
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