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

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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}