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.
Pattern Number | Number of Squares | 1st Difference | 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.
Pattern Number | Extra Bit | 1st Difference | 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
Therefore so far my formula is:3n2 + [extra bit]
I will now attempt to find the extra bit for this formula.
Pattern Number | Extra Bit | 1st Difference | 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]
Pattern Number | Extra Bit | 1st Difference | 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.
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