More on Figurate NumbersAuthor(s): Amit MozoomdarSource: Mathematics in School, Vol. 14, No. 2 (Mar., 1985), pp. 29-30Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30213974 .

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mTIC onn~rt t

by Amit Mozoomdar Four Dwellings Comprehensive School. Birmingham

After reading recent articles on the figurate numberst'2, i.e. the pentagonal, hexagonal and other "polygon numbers", I asked myself whether there was a simple way of deriving a formula for the nth term in the sequence of m-gon numbers. This is what I came up with:

Let PA' denote the nth m-gon number. For example:

P3 (triangular 1 3 6 10 15 21... numbers) are Pi (square 1 4 9 16 25 36... numbers) are Pi (pentagonal 1 5 12 22 35 51... numbers) are P6 (hexagonal 1 6 15 28 45 66... numbers) are

Now consider the differences between successive terms:

P3

4- n

P -

ps

np Pi-

-

115 21..

2/3

4 5 6

1 4 9 16 25 36...

\3/ \/ \ / \ / \ /

3 5 7 9 11

1 5 12 22 35 51...

4 7 10 13 16

1 6 15 28 45 66...

\/\6/\/\/ \ 5 6 13 17 21

The differences are in arithmetic progression, as men- tioned by Tourret', and the "differences between the differences" are given by m - 2, where the polygon has m sides. Therefore

P' = Pen-l

+ (m - 2)(n - 1) + 1 (1)

This is the equation relating the difference between suc- cessive m-gonal numbers to m and to n, the position in the sequence. After this one can eliminate

P"_ 1 by writing it in

terms of P"-2 and likewise by counting back along the sequence we can get to the first term P'

- 1. P' =

Pm-, + (m - 2)(n - 1) + 1

Rearrange:

P' = P~-,

+ n(m - 2) - (m - 3)

Count back:

Pm = 1 + (m - 2){n + (n - 1) + (n - 2) + ... + 2} - (m - 3)(n - 1) (2)

It is well-known that the sum of the integers from 1 to n is given by:

n(n + 1) Sn

n2

Substituting for this in (2) we get: p=1+(m

2n(n )

1)- (m -3)( n-1)

2 + (m - 2)(n2 + n- 2) - 2(m - 3)(n - 1) PA = n 2

m

=

(m- 2)n2 + n{(m - 2) - 2(m- 3)}+ 2- 2(m- 2) + 2(m- 3) n 2

(m- 2)n2 + n(m - 2 - 2m + 6)+ 2 - 2m + 4 + 2m - 6 P n 2

pm =

(m - 2)n2-(m 4) n- 2

It is easy to verify that this formula does in fact generate the nth term of any m-gonal number sequence. For m = 3 we get the triangle numbers

n2 -

n n(n + 1) P,3 , or more commonly TW 2 2

For m = 4 we get the squares

P4 = n2 For m = 5 we get the pentagonal numbers

3n2 - n

For m = 6 we get the hexagonal numbers

P6 = 2n2 _ n

Parabolas The formulae for the general terms in the sequences are not continuous functions because like all number-theoretic functions they apply only to positive integral values of the independent variable, but if we can ignore this for a

Mathematics in School, March 1985 29

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moment and plot a graph of P" against n we get a very nice infinite set of parabolas.

60 b to t

50

40

A.AA I

10 2

A12 3 4 5 6 7

The parabolas lead to an interesting consideration; for a fixed value of n (say the 5th term) how does the value of P" vary as the order (m) of the polygon number increases?

n Triangle Squares Pentagonal Hexagonal Differ- numbers numbers numbers ences

1 1 1 1 1 0

2 3 4 5 6 0

3 6 9 12 15 3

4] 10 16 22 28 6

5 15 25 35 45 10

6 21 36 51 66 15

Notice that the column of differences contains none other than our old friend, the sequence of triangle numbers. Thus:

Pn - Pn- = - 1

where T_

_, denotes the (n - 1)th triangle number.

Proof: (m - 2)n2 - (m - 4)n Pm =

n 2

P~" _ _ {(m - 2) - (m - 3)}n2 - {(m -

4) - (m - 5)}n

,2 pm p-1 _n2 - n

2

2

The reader can easily verify that the right-hand side of this identity is the formula for the (n - 1)" triangle number.

References

1. Tourret, A. V. (1981) More on Hexagonal Numbers, Mathematics in School, 10, 2.

2. Wyvill, R. (1983) The Multiplication Configurate, Mathematics in School, 12, 3.

"RK] [R PA. PA.

by Mike Rose Ilfracombe School and Community College, Devon

1 1 1 1 l 1 2 13 4 y 6

1 3 6 10 15 21

1 4 10 20 35 56

1 5 15 35 70

1 6 21 56 6

All mathematics teachers and hopefully most students of mathematics will be familiar with the triangular number array known as Pascal's triangle. The many properties of this pleasing array are well known but are often spoilt for pupils by the chore of having to draw out the array itself. Thus at its simplest level the work sheet given at the end of this article might provide (slower) pupils with an opportun- ity to succeed with the interesting mathematics.

The columns on the right hand side can be used variously to draw out the pattern of the sums of rows

e.g. 2 = 21 4 = 2 x 2 = 22 8=2x4=23

which most pupils can use to predict the sum of the 100th row etc. Colouring odd and even numbers in the main pattern of the array is an equally fruitful source of pleasure for pupils. Which row numbers will contain all odd num- bers for example. (See the sum of the rows for a clue.)

Pascal's triangle is but one of many patterns of this kind however and the following notes provide a simple classifi- cation of the various forms which can be experimented with. All the patterns are found to have a simple row total pattern which leads to some simple factorisation with numbers or alternatively with algebraic form patterns.

As it is useful for teachers and hopefully sixth-formers to know the number that comes in the middle of the bottom line, or anywhere else, I include the formula for work- ing out the number on the nth row in the rth position, r = 0, 1, 2,..., n which is called "X, for short.

30 Mathematics in School, March 1985

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