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# Growing Squares

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Introduction

John McLaughlin – [email protected]

Feel free to e-mail me on ways to improve this essay it would be a great help.

## Growing Squares

I have decided to find a formula to find the nth term.  To help me find the nth term I shall compose a table including all the results I know.

### 2nd Difference

1

1

4

4

2

5

8

3

13

4

12

4

25

The 2nd difference is constant; therefore the equations will be quadratic.  The general formula for a quadratic equation is an2 + bn +c.  The coefficient of n2 is half that of the second difference

Therefore so far my formula is:2n2 + [extra bit]

I will now attempt to find the extra bit for this formula.

### 2nd Difference

1

2

6

4

2

8

10

3

18

4

14

4

32

From my table of results I have found the formula to be 2n2 + 2n + 1

I will now check my formula by substituting a value from the table in to my formula:

E.g. n = 2

Un = 2 (2) 2 – 2 (2) + 1 = 8.

For Diagrams 1 – 4 I can see a pattern with square numbers.

Middle

2 + bn +c.  The coefficient of n2 is half that of the second difference

Therefore so far my formula is:3n2 + [extra bit]

I will now attempt to find the extra bit for this formula.

### 2nd Difference

1

0

6

6

2

6

12

3

18

6

18

4

36

From my table of results I have found the formula to be3n2 + 3n + 1

I will now check my formula by substituting a value from the table in to my formula:

E.g. n=3

Un=3(3) 2 -3(3) + 1 = 19.

For my table of results I can see that my formula is correct after verification.

I have chosen to justify my formula using 3 simultaneous equations rather that geometrically like the previous investigation.

The standard quadratic formula » U = an2 + bn + c

I will now substitute values for U and n

7 = a (2) 2 + 2b + c                        (1)

19 = a (3) 2 + 3b + c                        (2)

37 = a (4) 2 + 4b + c                        (3)

If we then subtract equation (2) from (1) we get

12 = 5a +b                                (4)

When we do the same between (3) and (2)

18 = 7a + b                                (5)

(5) – (4) = 6 = 2a

Therefore a = 3

To work out b we must insert the value of a in to equation (4)

(4) » 12 = 5 (3)

Conclusion

Therefore so far my formula is:4/3n2 - 2n² [extra bit]

### 2nd Difference

1

12/3

2

41/3

3

7

4

91/3

5

121/3

Therefore my formula is:

4/3n³ – 2n² + 8/6n – 1

I will test this formula out on a pattern previously drawn out and recorded.

Un = 4/3n³ – 2n² + 8/6n – 1

36 – 18 + 4 – 1

= 25

This is correct

I will now and justify the formula by using simultaneous equations.

Let C= An³ - bn² +cn – d

When n = 1, c = 1.  Therefore:

1= a +b + c + d                                         (1)

When n = 2, c = 7.  Therefore:

7 = 8a + 4b + 2c + d                                (2)

When n = 3, c = 25.  Therefore:

25 = 27a +9b + 3c + d                                (3)

When n = 4, c = 65.  Therefore:

63 = 64a + 16b + 4c + d                        (4)

If we then subtract equation (2) from equation (1) we get

6 = 7a + 3b + c                                        (5)

If we do the same again to (3) and (2), we get

18 = 19a + 5b + c                                (6)

Then to (4) and (3)

38 = 37a + 7b + c                                (7)

Then to (6) and (5)

12 = 12a + 2b                                        (8)

Then again to (7) – (6)

20 = 18a + 2b                                        (9)

And finally to (9) and (8)

A = 8/6 = 4/3

Substitute into 8

12 = 16 + 2b

-2 = b

Substitute into 5

6 = 9 1/3 – 6 + c

2 2/3 = c = 8/3

Substitute into 1

1 = 4/3 + (-2) + 8/3 + d

-1 = d

Therefore my formula is as follows;

C=4/3n³ – 2n² + 8/6n – 1

This is correct.

John McLaughlin – [email protected]

Feel free to e-mail me on ways to improve this essay it would be a great help.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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