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## 'The Root of the Problem' printed from http://nrich.maths.org/

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.

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}