You may also like

problem icon

Telescoping Series

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

problem icon

The Kth Sum of N Numbers

Yatir from Israel describes his method for summing a series of triangle numbers.

problem icon

Binomial Coefficients

An introduction to the binomial coefficient, and exploration of some of the formulae it satisfies.

The Harmonic Triangle and Pascal's Triangle

Stage: 5
Article by Toni Beardon

We shall explore the similarities between the harmonic triangle, which you see below, and Pascal's triangle. You may like to start by trying this Harmonic Triangle problem.
The rule for generating the harmonic triangle is that you add two consecutive entries to give the entry between them in the row above. Equivalently, from any term, to get the next term, subtract that term from the corresponding term on the row above. It is still possible to work downwards row by row because the entry at the left hand end of the $n$th row is ${1\over n}$.

For example the third row of the harmonic triangle is: $${1\over 3},\ {1\over 6},\ {1\over 3}$$ and the fourth row is: $${1\over 4},\ {1\over 12},\ {1\over 12}, \ {1\over 4}$$ which is given by $${1\over 4},\ \ \left[{1\over 3}-{1\over 4}\right] ,\ \ \left[{1\over 6}-{1\over 12}\right],\ \ \left[{1\over 3}-{1\over 12}\right]$$
 \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} \\ & & & & & ... & & & & & \end{array}

So is it possible to continue generating the harmonic triangle indefinitely using this rule and will all the terms be fractions with unit numerators?

The entries in the harmonic triangle are related by a similar rule to the entries in Pascal's triangle and both sets of entries involve the binomial coefficients. We denote the $r$th entry in the $n$th row of Pascal triangle by the binomial coefficient: $${n-1\choose r-1}= {(n-1)!\over (r-1)!(n-r)!}.$$ For example the sixth row is $${5 \choose 0} = 1 , {5 \choose 1}= 5, {5\choose 2}=10, {5\choose 3}= 10, {5\choose 4}= 5, {5\choose 5} =1 $$ \begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1& & 1 & & & & \\ & & & 1 & & 2 & & 1 & & & \\ & & 1 & & 3 & & 3 & & 1 & & \\ & 1& & 4 & & 6 & & 4 & & 1 & \\ 1& & 5& & 10 & & 10 & & 5 & & 1 \\ & & & & & ... & & & & & \end{array}

The rule for generating Pascal's triangle is that you add two consecutive entries to give the entry between them in the row below. This rule is given by $${n-1\choose r-1} + {n-1 \choose r} = {n\choose r}.$$ This is a well known result, it can be proved by simple algebra and the proof is given in many textbooks.

For the harmonic triangle the $r$th entry in the $n$th row is given by: $$H(n, r) = {(r-1)!\over n(n-1)(n-2)...(n-r+1)}= {(r-1)!(n-r)!\over n!}= {1\over n{n-1\choose r-1}}$$ Knowing that the binomial coefficient ${n-1\choose r-1}$ is a whole number it is clear from this formula that all the fractions in the harmonic triangle will have $1$ as numerator.

It remains to prove that the rule for generating the harmonic triangle works in general for all $n$ and $r$ and then we shall have established that the harmonic triangle can be extended to $n$ rows for any $n$ by using the given formula for the $r$th entry in the $n$th row.

The harmonic triangle rule is given by the formula: $$H(n, r) + H(n, r+1) = H(n-1, r)$$ or equivalently, as described above, $$H(n, r+1) = H(n-1, r)-H(n, r).$$ The proof is left to the reader. All you need to do is substitute the formulae in terms of the factorials into the right hand side of this identity and simplify it to get the left hand side. If you want to check that you have managed to do it correctly you can click on Solution above.

The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
Copyright © 1997 - 2012. University of Cambridge. All rights reserved.
NRICH is part of the family of activities in the Millennium Mathematics Project.