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## 'Weekly Challenge 18: the Root of the Problem' printed from http://nrich.maths.org/

This challenge previously feature on NRICH.
Correct solutions were recieved from Charlene from Brunei, Kiang
from Singapore, Andre from Bucharest and Jing of Madras College.
Well done to all of you. Charlene's solution is given below. Not as
hard as it at first looks! The moral is not to be put off by
appearances.
A great discussion took place
concerning this challenge on Ask
NRICH.

The solution makes use of
'rationalising the denominator' in which the denominator of each
term is converted to an integer by multiplication with a factor
chosen so as to use the expression
$$

(x-y)(x+y) = x^2-y^2

$$

In detail, the numerator and denominator of the terms can be
multiplied to give a more convenient value as follows:

\begin{eqnarray}&&\frac{1 \times(\sqrt{1} -
\sqrt{2})}{(\sqrt{1} + \sqrt{2})(\sqrt{1} - \sqrt{2})} + \frac{1
\times (\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3})(\sqrt{2} -
\sqrt{3})} + \dots + \frac{1 \times (\sqrt{99} -
\sqrt{100})}{(\sqrt{99} + \sqrt{100})(\sqrt{99} - \sqrt{100})}\\
&=& \frac{(\sqrt{1} - \sqrt{2})}{-1} + \frac{(\sqrt{2} -
\sqrt{3})}{-1} + \dots + \frac{(\sqrt{99} - \sqrt{100})}{-1}\\
&=& (-\sqrt{1} + \sqrt{2}) + (-\sqrt{2} + \sqrt{3}) + \dots
+ (-\sqrt{99} + \sqrt{100}) \\ &=& -\sqrt{1} + \sqrt{100}\\
&=& -1 + 10 \\ &=& 9 \end{eqnarray}