# The Root of the Problem

Can you get to the root of this problem about surds?

## Problem

Find the sum of $$\frac{1}{\sqrt{1}+ \sqrt{2}}+ \frac{1}{\sqrt{2}+ \sqrt{3}} + \text{and so on up to}+\frac{1}{ \sqrt {99}+ \sqrt{100}}.$$

Can you invent any similar sums which have integer answers?

Did you know ... ?

Whilst this series can be summed using elementary methods mathematicians devise various ways in which the sums of series can be analysed. These are explored in greater detail in university analysis courses.

Whilst this series can be summed using elementary methods mathematicians devise various ways in which the sums of series can be analysed. These are explored in greater detail in university analysis courses.

## Student Solutions

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}