Mock AIME 3 Pre 2005 Problems/Problem 6

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 $\sqrt{n+\sqrt{n^2-1}}=a+b\sqrt{c}$ $\sqrt{n-\sqrt{n^2-1}}=a-b\sqrt{c}$ for some $a$ , $b$ , $c$ .

Squaring the first equation yields $n+\sqrt{n^2-1}=a^2+b^2c+2ab\sqrt{c}$ which gives the system of equations $n=a^2+b^2c$ $\sqrt{n^2-1}=2ab\sqrt{c}$

Then we have $\frac{1}{a+b\sqrt{c}}=a-b\sqrt{c}$ $a^2-b^2c=1$

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by Problem 5	Followed by Problem 7
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Mock AIME 3 Pre 2005 Problems/Problem 6

Contents

Problem

Solution

Solution 2

See Also