Subject:

Sequences (Mathematics)
(Tests, problems and exercises)

Pub Date:

06/22/2011

Publication:

Name: Australian Mathematics Teacher Publisher: The Australian Association of Mathematics Teachers, Inc. Audience: Academic Format: Magazine/Journal Subject: Education; Mathematics Copyright: COPYRIGHT 2011 The Australian Association of Mathematics Teachers, Inc. ISSN: 0045-0685

Issue:

Date: Summer, 2011 Source Volume: 67 Source Issue: 2

Geographic:

Geographic Scope: Australia Geographic Code: 8AUST Australia

Accession Number:

261951287

Full Text:

Number patterns

Consider the counting numbers

1, 2, 3, 4, 5, ...

which shall be labelled C(n). If the odd counting numbers are labelled O(n) and the even counting numbers are labelled E(n), it follows that C(n) = O(n) [union] E(n). Here your students are being introduced to discrete function notation.

Euclid wrote his book on the elements of geometry in the 3rd century BC, but it was not until about AD 100 that Nichomachus, a Greek living near Jerusalem, produced his "Introduction to Arithmetic" which collected in one publication the connection between geometry and arithmetic. Nichomachus did this by drawing number patterns as geometrical shapes. For example, the triangular numbers 1, 3, 6, and 10 are illustrated in Figure 1.

[FIGURE 1 OMITTED]

Using the sum of an arithmetic progression with common difference 1, it can be shown that the nth triangular number is given by

T (n) = n(n + 1)/2

Ask your students to draw the square numbers and write down the formula for them [S(n) = [n.sup.2]]. Next ask them to draw the oblong numbers OB1(n) formed from rectangles with the number of columns (n) one less than the number of rows (n + 1).When they write down the formula for OB1(n), they should be able to see the connection with T(n).

There is also a connection between T(n) and S(n), and this can be discovered by partitioning any square diagram appropriately. Figure 2 shows how it is done for S(4).

[FIGURE 2 OMITTED]

The square and oblong numbers can be considered as special cases of the more general rectangular numbers Rk(n) where k denotes the integer difference between the number of rows and columns in the rectangular array. Thus R0(n) = S(n) and R1(n) = OB1(n). Ask your students to find formulae for R2(n) and R3(n).

There is also a connection between the square numbers and the odd counting numbers, as illustrated in Figure 3.

[FIGURE 3 OMITTED]

See if your students can discover this connection from the diagram.

Following directly on from the triangular and square numbers are the pentagonal numbers P(n), and these will now be developed geometrically. The first pentagonal number P(1) is 1 and the second pentagonal number P(2) is clearly 5. From the diagram in Figure 4 it is seen that the sides adjacent to one vertex and the diagonals from that vertex are extended to form the next similar pentagon. The number of dots on each side of this extended pentagon are increased by 1. Therefore the next two pentagonal numbers are P(3) = 12 and P(4) = 22.

[FIGURE 4 OMITTED]

Your students should now be able to discover for themselves the next two pentagonal numbers and maybe more. After this they should go on to establish the hexagonal numbers HX(n) and the heptagonal numbers HP(n). The following table can then be constructed.

On completion of this table your students should note the patterns in the various columns.

Nichomachus also considered the connections between numbers and three-dimensional geometrical patterns, such as the cubic numbers and the pyramidal numbers. The latter had practical use in the 18th and 19th centuries in counting the number of cannonballs in a pyramidal stack.

The square numbers S(n) can also be used to help your students discover other simple patterns and develop their mathematical skills with numbers. Consider the difference between two squares written algebraically as

D(a,b) = [a.sub.2] - [b.sub.2] = (a - b)(a + b)

Note that two-variable functional notation has now been introduced.

First of all write a = b + 1, then clearly a - b = 1 and a + b = 2b + 1. Thus

D(1,b) = 2b + 1

and therefore, as b progresses through the counting numbers, the odd numbers are generated. This, of course, can be seen from Figure 3 which was first illustrated by Nichomachus.

Next write a = b + 2 and then

D(2, b) = 2(2b + 2) = 4b + 4

which is simply an arithmetic progression with common difference 4 when b progresses through the C(n). With a = b + 3 then

D(3, b) = 3(2b + 3)

an arithmetic progression with common difference 6. Thus

D(k, b) = k(2b + k)

for k = 1, 2, 3, are all arithmetic progressions with common difference 2k. Therefore your students have now discovered their first patterns with two counting elements k and b. Ask them to draw up a table for D(k, b) with k-values as the row headings and b-values as the column headings. Look at the patterns in the table.

Finally, consider the reverse process: namely, given a number N find all the different possible Ds that evaluate to N. For example, suppose that N = 45. The possible solutions can be obtained by considering the various D(k,b) patterns.

Taking k = 1 yields 45 = 2b + 1, and so b = 22, k = 23 and D(23, 22) = 45.

Taking k = 2 yields no solution, since b cannot be fractional. The same holds for all k = E(n). Taking k = 3 yields D(9, 6), while k = 5 produces D(7, 2). That is all that can be found, since taking k = 7 produces b as a negative number. Ask your students to find a connection between these three solutions and the prime factors of 45.

Give your students other values of N to try. Some may have no solutions (N = 2) or only one solution (N = 7). Find the condition that there is only one solution for a particular value of N. Have a look at N = 945, and explain why there are only eight solutions. This last problem was suggested to me by Bruce Cameron. Happy discoveries!

Consider the counting numbers

1, 2, 3, 4, 5, ...

which shall be labelled C(n). If the odd counting numbers are labelled O(n) and the even counting numbers are labelled E(n), it follows that C(n) = O(n) [union] E(n). Here your students are being introduced to discrete function notation.

Euclid wrote his book on the elements of geometry in the 3rd century BC, but it was not until about AD 100 that Nichomachus, a Greek living near Jerusalem, produced his "Introduction to Arithmetic" which collected in one publication the connection between geometry and arithmetic. Nichomachus did this by drawing number patterns as geometrical shapes. For example, the triangular numbers 1, 3, 6, and 10 are illustrated in Figure 1.

[FIGURE 1 OMITTED]

Using the sum of an arithmetic progression with common difference 1, it can be shown that the nth triangular number is given by

T (n) = n(n + 1)/2

Ask your students to draw the square numbers and write down the formula for them [S(n) = [n.sup.2]]. Next ask them to draw the oblong numbers OB1(n) formed from rectangles with the number of columns (n) one less than the number of rows (n + 1).When they write down the formula for OB1(n), they should be able to see the connection with T(n).

There is also a connection between T(n) and S(n), and this can be discovered by partitioning any square diagram appropriately. Figure 2 shows how it is done for S(4).

[FIGURE 2 OMITTED]

The square and oblong numbers can be considered as special cases of the more general rectangular numbers Rk(n) where k denotes the integer difference between the number of rows and columns in the rectangular array. Thus R0(n) = S(n) and R1(n) = OB1(n). Ask your students to find formulae for R2(n) and R3(n).

There is also a connection between the square numbers and the odd counting numbers, as illustrated in Figure 3.

[FIGURE 3 OMITTED]

See if your students can discover this connection from the diagram.

Following directly on from the triangular and square numbers are the pentagonal numbers P(n), and these will now be developed geometrically. The first pentagonal number P(1) is 1 and the second pentagonal number P(2) is clearly 5. From the diagram in Figure 4 it is seen that the sides adjacent to one vertex and the diagonals from that vertex are extended to form the next similar pentagon. The number of dots on each side of this extended pentagon are increased by 1. Therefore the next two pentagonal numbers are P(3) = 12 and P(4) = 22.

[FIGURE 4 OMITTED]

Your students should now be able to discover for themselves the next two pentagonal numbers and maybe more. After this they should go on to establish the hexagonal numbers HX(n) and the heptagonal numbers HP(n). The following table can then be constructed.

On completion of this table your students should note the patterns in the various columns.

Nichomachus also considered the connections between numbers and three-dimensional geometrical patterns, such as the cubic numbers and the pyramidal numbers. The latter had practical use in the 18th and 19th centuries in counting the number of cannonballs in a pyramidal stack.

The square numbers S(n) can also be used to help your students discover other simple patterns and develop their mathematical skills with numbers. Consider the difference between two squares written algebraically as

D(a,b) = [a.sub.2] - [b.sub.2] = (a - b)(a + b)

Note that two-variable functional notation has now been introduced.

First of all write a = b + 1, then clearly a - b = 1 and a + b = 2b + 1. Thus

D(1,b) = 2b + 1

and therefore, as b progresses through the counting numbers, the odd numbers are generated. This, of course, can be seen from Figure 3 which was first illustrated by Nichomachus.

Next write a = b + 2 and then

D(2, b) = 2(2b + 2) = 4b + 4

which is simply an arithmetic progression with common difference 4 when b progresses through the C(n). With a = b + 3 then

D(3, b) = 3(2b + 3)

an arithmetic progression with common difference 6. Thus

D(k, b) = k(2b + k)

for k = 1, 2, 3, are all arithmetic progressions with common difference 2k. Therefore your students have now discovered their first patterns with two counting elements k and b. Ask them to draw up a table for D(k, b) with k-values as the row headings and b-values as the column headings. Look at the patterns in the table.

Finally, consider the reverse process: namely, given a number N find all the different possible Ds that evaluate to N. For example, suppose that N = 45. The possible solutions can be obtained by considering the various D(k,b) patterns.

Taking k = 1 yields 45 = 2b + 1, and so b = 22, k = 23 and D(23, 22) = 45.

Taking k = 2 yields no solution, since b cannot be fractional. The same holds for all k = E(n). Taking k = 3 yields D(9, 6), while k = 5 produces D(7, 2). That is all that can be found, since taking k = 7 produces b as a negative number. Ask your students to find a connection between these three solutions and the prime factors of 45.

Give your students other values of N to try. Some may have no solutions (N = 2) or only one solution (N = 7). Find the condition that there is only one solution for a particular value of N. Have a look at N = 945, and explain why there are only eight solutions. This last problem was suggested to me by Bruce Cameron. Happy discoveries!

Table 1 n 2 3 4 5 6 T(n) 3 6 10 15 21 S(n) 4 9 16 25 36 P(n) 5 12 22 35 51 HX(n) 6 15 28 45 66 HP(n) 7 18 34 55 81 OC(n) 8 21 40 65 96

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