AoPSWiki

Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!

Personal tools

Search

Toolbox

2006 Cyprus MO/Lyceum/Problem 15

From AoPSWiki

Problem

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

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

Solution

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

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

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

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_Cyprus_MO/Lyceum/Problem_15"

Category: Introductory Algebra Problems

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us