IMC 2017 Q24
24. The \(n\)th term in a certain sequence is calculated by multiplying together all the numbers \(\sqrt{1 + \dfrac{1}{k}}\), where \(k\) takes all the integer values from 2 to \(n + 1\) inclusive. For example, the third term in the sequence is \(\sqrt{1 + \dfrac{1}{2}} \times \sqrt{1 + \dfrac{1}{3}} \times \sqrt{1 + \dfrac{1}{4}}\).
Which is the smallest value of \(n\) for which the \(n\)th term of the sequence is an integer?
- A 3
- B 5
- C 6
- D 7
- E more than 7
Intermediate Mathematical Challenge 2017, Question 24. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 6
The product may be written \(\sqrt{\frac{3}{2}} \times \sqrt{\frac{4}{3}} \times \sqrt{\frac{5}{4}} \times \sqrt{\frac{6}{5}} \times \ldots = \frac{\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{4}}{\sqrt{3}} \times \frac{\sqrt{5}}{\sqrt{4}} \times \frac{\sqrt{6}}{\sqrt{5}} \times \ldots\).
Notice that the numerator of each fraction is cancelled out by the denominator of the following fraction and the only terms which are not cancelled are the denominator of the first fraction and the numerator of the last fraction. So the \(n\)th term of the sequence is \(\frac{\sqrt{n + 2}}{\sqrt{2}} = \sqrt{\frac{n + 2}{2}}\). As \(n \geqslant 1\) the product is not equal to 1.
The product increases with \(n\). So the next possible integer value to consider is 2 and this does occur when \(n = 6\) as \(\sqrt{\frac{6 + 2}{2}} = \sqrt{4} = 2\).
So the smallest number of terms required for the product to be an integer is 6.
Solution © UK Mathematics Trust, from the IMC 2017 solutions. Longer versions with investigations: Solutions and investigations (PDF).