# 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 $$n=a^2+b^2c$$ $$\sqrt{n^2-1}=2ab\sqrt{c}$$ calling them equations $A$ and $B$, respectively.

Also we have $$\frac{1}{\sqrt{n+\sqrt{n^2-1}}}=\sqrt{n-\sqrt{n^2-1}}$$ $$\frac{1}{a+b\sqrt{c}}=a-b\sqrt{c}$$ $$a^2-b^2c=1$$ which obtains equation $C$.

Adding equations $A$ and $C$ yields $$2a^2=n+1$$ $$a=\sqrt{\frac{n+1}{2}}$$ Squaring equation $B$ and substituting yields $$4a^2b^2c=n^2-1$$ $$2\cdot(n+1)\cdot b^2c=n^2-1$$ $$b^2c=\frac{(n-1)(n+1)}{2\cdot(n+1)}$$ $$b\sqrt{c}=\sqrt{\frac{n-1}{2}}$$

Thus we obtain the telescoping series \begin{align*} S &= \sum_{n=1}^{9800}a-b\sqrt{c} \\ &= \sum_{n=1}^{9800}\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}} \\ \end{align*}

Simplifying the sum we are left with \begin{align*} S &= -\sqrt{\frac{1}{2}}+\sqrt{\frac{9800}{2}}+\sqrt{\frac{9801}{2}} \\ &= -\frac{\sqrt{2}}{2}+\frac{99\sqrt{2}}{2}+70 \\ &= 70+49\sqrt{2} \\ \end{align*}

Thus $p+q+r=70+49+2=\boxed{121}$.

~ Nafer