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This is the start of the harmonic triangle:

\begin{array}{ccccccccccc} & & & & &\frac{1}{1} & & & & & \\ & & & & \frac{1}{2} & & \frac{1}{2} & & & & \\ & & & \frac{1}{3} & &\frac{1}{6} & & \frac{1}{3} & & & \\ & & \frac{1}{4} & &\frac{1}{12} & & \frac{1}{12} & & \frac{1}{4} & & \\ & \frac{1}{5} & & \frac{1}{20} & & \frac{1}{30} & & \frac{1}{20} & & \frac{1}{5} & \\ \frac{1}{6} & & \frac{1}{30} & & \frac{1}{60} & & \frac{1}{60} & & \frac{1}{30} & & \frac{1}{6}\\ & & & & & \ldots& & & & & \end{array}

Each fraction is equal to the sum of the two fractions below it.

Look at the triangle above and check that the rule really does work.

Can you work out the next two rows?

The $n$th row starts with the fraction $\frac{1}{n}$.We can continue the first diagonal ($\frac{1}{1}$, $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, and so on) using this rule.

Take a look at the second diagonal: ($\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{12}$, $\frac{1}{20}$, and so on).

What do you notice about the numerators and denominators of these fractions?

Take a look at the second diagonal: ($\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{12}$, $\frac{1}{20}$, and so on).

What do you notice about the numerators and denominators of these fractions?

Can you prove the pattern will continue?

What about the third and fourth diagonals?

Click here for a poster of this problem.