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Maths Coursework: Consecutive Numbers

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Introduction

MARTIN GALLAGHER

Maths Coursework: Consecutive Numbers

Introduction

        In this coursework I am trying to find a common pattern in consecutive numbers, and then solve the pattern with algebra.

Problem 1:

        Write down three consecutive numbers. Square the middle number; multiply the first number by the third. Compare your answers, what do you notice?

Problem 2:

        Write down two consecutive numbers. Square both of the numbers and find the difference between the two squares, what do you notice?

Problem 3:

        Write down five consecutive numbers. Square the middle number; multiply the first number by the fifth. Compare your answers, what do you notice?

Problem 4:

        Write down three consecutive numbers but this time with a gap between them, like this: 1, 3, 5. Square the middle number; multiply the first number by the third. Compare your answers, what do you notice?

Problem 1

Chosen consecutive numbers

Middle number squared

First number multiplied by the last.

Difference

9, 10, 11

5, 6, 7

7, 8, 9

10, 11, 12

18, 19, 20

100

36

64

121

361

99

35

63

120

360

1

1

1

1

1

In the

...read more.

Middle

20

21

400

441

41

15

16

225

256

31

65

66

4225

4356

131

128

129

16384

16641

257

From this table you can see that “Differences between squared numbers” column is the same number I get when I add the two numbers in the “Chosen numbers” column.

From this pattern I will try to find a rule in algebra.

Formula

n, n + 1

= n² - (n + 1)²

= n² - (n + 1)( n + 1)

= n² - n² + n + n +1

= n² - n² + 2n + 1

= 2n + 1

Test the formula

To test the above algebra and to prove I am correct I must now test the formula. My formula for this problem is “2n + 1”, I must replace “n” with one of the consecutive numbers I have chosen. For example: if n = 5, the sum will look like this: 2 x 5 + 1 = 11.

I will test my formula in this table:

Chosen consecutive number

After equation (2n + 1)

5

20

15

65

128

11

41

31

131

257

As the above table shows my formula is correct. The answers I get from this table match my earlier table correctly.

I have now done “Problem 2” but there is no clear pattern I can see, so I will find more rules for different consecutive patterns.

Problem 3

Chosen consecutive numbers

Middle number squared

First number multiplied by the fifth

Difference

1, 2, 3, 4, 5

20, 21, 22, 23, 34

34, 35, 36, 37, 38

45, 46, 47, 48, 49

12, 13, 14, 15, 16

9

484

1296

2209

196

5

480

1292

2205

192

4

4

4

4

4

...read more.

Conclusion

Formula

n, n + 2, n + 4

= n(n + 4) - (n + 2)²

= n² + 4n - (n + 2)(n + 2)

= n² + 2n + 2n + 4 - n² + 4n

= n² + 4n + 4 - n² + 4n

= 4

Testing the formula

When n = 8

= 8(8 + 4) - (8 + 2)²

= 8² + 4 x 8 - (8 + 2)(8 +2)

= 8² + 2 x 8 + 2 x 8 + 4 - 8² + 4 x 8

= 8² + 4 x 8 + 4 - 8² + 4 x 8

= 4

I can now see it is highly likely my prediction was correct and the pattern being: (Gap between consecutive numbers)².

__________________________________________________________________________

Working out the formula

The final formula for working out the pattern in consecutive numbers is shown below:

n, n + g, n + 2g

= n(n +2g) + (n + g)²

= n(n +2g) + (n + g) (n + g)

= (n² + 2gn) + (n² + gn + gn +g²)

= (n² + 2gn + g²) – (n² + 2gn)

= g²

        With the above formula I can workout any consecutive number sum, I replace “n” with the number(s) used and change “g” with the amount of the gap between each number.

      E.g. if the consecutive numbers chose where: 2, 4, 6 the “g” in the formula would be replaced with 2, because that’s the size of the gap between numbers in this case.

...read more.

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