2006 Cyprus MO/Lyceum/Problem 15

Problem

The expression $\frac{1}{2+\sqrt7} + \frac{1}{\sqrt7+\sqrt{10}}+ \frac{1}{\sqrt{10}+\sqrt{13}} + \frac{1}{\sqrt{13}+4}$ equals

$\mathrm{(A)}\ \frac{3}{4}\qquad\mathrm{(B)}\ \frac{3}{2}\qquad\mathrm{(C)}\ \frac{2}{5}\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{2}{3}$

Solution

Multiply all of the terms by their complex conjugates to simplify:

$\frac{1}{\sqrt{7} + \sqrt{4}} \cdot \left(\frac{\sqrt{7}-\sqrt{4}}{\sqrt{7}-\sqrt{4}}\right) + \ldots + \frac{1}{\sqrt{16} + \sqrt{13}} \cdot \left(\frac{\sqrt{16}-\sqrt{13}}{\sqrt{16}-\sqrt{13}}\right)$ $= \frac{\sqrt{7} - \sqrt{4}}{3} + \frac{\sqrt{10} - \sqrt{7}}{3} + \frac{\sqrt{13} - \sqrt{10}}{3} + \frac{\sqrt{16} - \sqrt{13}}{3}$

This telescopes to $\frac{\sqrt{16} - \sqrt{4}}{3} = \frac{2}{3} \Longrightarrow \mathrm{(E)}$ .

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 14	Followed by Problem 16
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2006 Cyprus MO/Lyceum/Problem 15

Problem

Solution

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