# The A to Z of mathematics a basic guide - Sidebotham T.H.

ISBN 0-471-15045-2

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1 1=2Â°

1 1 1 + 1 = 2 = 21

1 2 1 1 + 2 + 1 = 4 = 22

1 3 3 1 l + 3 + 3 + l= 8=23

1 4 6 4 1 1+4+6+4+1 = 16 = 24

1 5 10 10 5 1 1 + 5 + 10 + 10 + 5 + 1 = 32 = 25

PATTERNS 319

The triangle drawn here contains six rows, but more rows can be added using the patterns of numbers. One of the diagonal rows is the set of counting numbers 1, 2, 3, 4, 5,..., which are also called natural numbers. Below the natural numbers is a diagonal of triangle numbers, 1, 3, 6, 10, â€” Adding the numbers in each row give powers of 2, which are 1, 4, 9, 16, 32, â€”

Each number in the table can be expressed as a combination nCr, which is also known as a selection. For example, consider the row of numbers 1, 5,10,10,5, and 1:

5C0 = 1, 5Q = 5, 5C2 = 10, 5C3 = 10, 5C4 = 5, 5C5 = 1

The next row of Pascalâ€™s triangle is:

6Co = 1, 6C, = 6, 6C2 = 15, 6C3 = 20, 6C4 = 15, 6C5 = 6, = 1

More information on Pascalâ€™s triangle is given in the entry Coefficient.

References: Coefficient, Combinations, Natural Numbers, Powers, Triangle Numbers. PATTERNS

A set of numbers forms a pattern when it obeys a given rule. A pattern may also exist in a set of drawings or designs, which may also be represented by a given rule. In this entry the aim is to find the rule, or formula, which describes a pattern of numbers or drawings. The rule will be given in terms of n, and the pattern of numbers is obtained by substituting the numbers n = 1, 2, 3, 4,... one at a time into the rule.

Example 1. Harry is a postman and he is delivering letters to the houses with even numbers, which are on one side of the street. The numbers of the houses are 2, 4, 6, 8, â€” What is the number of the 20th house, and what is the formula that describes this pattern of house numbers?

Solution. The formula will involve a variable, which we call n, and we can introduce the variable n in a table the following way:

Position of house by counting 1st 2nd 3rd 4th 20th nth

Number on the house 2 4 6 8

The variable n is introduced as the nth house in the street of even numbers, and the formula we require is the house number of the nth house. The table contains a pattern that we need to identify. By studying the pattern of numbers in the table we can see that the 5th house is numbered 10, and the 6th house is numbered 12. So the 20th house is numbered 40. The pattern is to multiply n by 2, which equals 2n, to obtain the number of the nth house. The rule is: House numbers = 2n. The pattern of house numbers 2, 4, 6, 8,... is obtained by substituting n = 1, 2, 3, 4,... in turn into the rale House numbers = 2n.

320 PATTERNS

Example 2. What is the formula for the house numbers on the opposite side of the street, which is the set of odd numbers 1, 3, 5, 7,... ?

Solution. An odd number is one more than an even number, or one less. So the formula is either 2n + 1 or 2n â€” 1. We can check which of these two rules is correct by substituting for n the numbers 1, 2, 3, 4, â€”

â™¦ First rule: when n = 1, 2n + 1 = 3.

â™¦ Second rule: when w = 1,2w â€” 1 = 1.

The second rule is the correct formula, because the first house number is 1, and not 3. The formula for the house numbers 1, 3, 5,7,... is 2n â€” 1, where n is 1, 2, 3,4, â€”

The following table gives rules, in terms of ft, for three well-known patterns of

numbers:

Square numbers 1 4 9 16 25 /72

Triangle numbers 1 3 6 10 15 4/?(/?+ 1)

Cube numbers 1 8 27 64 125 /73

There is not a rule for the set of prime numbers 2, 3, 5, 7, 11, 13, 17, â€” Check in the entry Difference Tables for more assistance in finding the rules for patterns of numbers.

The following examples give patterns of figures which will be reduced to patterns of numbers to obtain the rules.

Example 3. Isabel sells square swimming pools, and is interested in knowing how many tiles are needed to tile around the pools. Each tile is 1 meter square. How many tiles are needed for a pool that is n meters square?

Solution. Isabel makes sketches of pools of increasing sizes and counts the tiles needed to fit around them (see figure a). She enters the data into a table in order to recognize the pattern more easily:

(a)

Length of side of pool in meters 12 3 4

Number of tiles to go around pool 8 12 16 20

n

4/7 + 4

PENTAGON 321

We will study the sketch of (say) the third drawing. There are 5 tiles along the top and 5 along the bottom. The 5 is made up of (3 + 2), where 3 meters is the length of the pool. So for a pool of length n meters, along the top and the bottom there are (n + 2) + (n T 2) = 2n + 4 tiles. Going back to the pool of length 3 meters, there are 3 tiles along the two remaining sides, where 3 meters is the length of the pool. So for a pool of length n meters there is an extra n + n = 2n tiles for the two remaining sides. For a square pool of length n meters there are 2n + 4 + 2n = 4n + 4 tiles needed. The formula can be checked by substituting, in turn, n = 1, 2, 3, 4,... into the formula 4n + 4. For a pool of side n meters, (4n + 4) tiles are needed to tile around it.

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