You may also like

problem icon

Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

problem icon

Rudolff's Problem

A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?

problem icon

Square Mean

Is the mean of the squares of two numbers greater than, or less than, the square of their means?


Age 14 to 16 Challenge Level:

Sunil sent us his work on this problem:

When I tried this out, no matter what numbers I started with, I always seemed to find that the $6$th term was the same as the $1$st term, and the $7$th the same as the $2$nd, so it kept repeating. I used a spreadsheet to help me do the calculations! Then I used algebra to try to explain why it worked.

The first term is $a_1$.

The second term is $a_2$.

The third term is $\frac{1+a_2}{a_1}$.

The fourth term is $\frac{1+a_1+a_2}{a_1 a_2}$.

The fifth term is $\left(\frac{1+a_1+a_2+a_1 a_2}{a_1 a_2}\right)\times\left(\frac {a_1}{1+a_2}\right)=\left(\frac{(1+a_1)(1+a_2)}{a_1 a_2}\right)\times\left( \frac{a_1}{1+a_2}\right)=\frac{1+a_1}{a_2}$.

The sixth term is $\left(\frac{1+a_1+a_2}{a_2}\right)\times\left(\frac{a_1 a_2} {1+a_1+a_2}\right)=a_1$.

The seventh term is $\left(a_1+1\right)\times\left(\frac{a_2}{1+a_1}\right)=a_2$.

This explains why the pattern always works, no matter what you start with! (As long as the bottom is never $0$, so we can't have $a_1=0$, or $a_2=0$, or $a_1=-1$, or $a_2=-1$, or $a_1+a_2=-1$.)