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

Revision as of 11:47, 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 $\begin{align*} S &= \sum_{n=1}^{9800}\frac{1}{\sqrt{n+\sqrt{n^2-1}}} \\ &= \sum_{n=1}^{9800}\frac{\sqrt{n+\sqrt{n^2-1}}}{n+\sqrt{n^2-1}} \\ \end{align*}$

@@ Line 22: / Line 22: @@
 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*}</cmath>

Mock AIME 3 Pre 2005 (Problems, Source)
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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"

Revision as of 11:47, 28 November 2019

Contents

Problem

Solution

Solution 2

See Also