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Introduction

YETOMIWA ARAYOMI 10A 1a) The numbers 5,12,13 satisfy the condition

a2 + b2 = c2

52 + 122 = 132

52  = 5 x 5 = 25

122 = 12 x 12 = 144

132 = 13  x 13 = 169

and so                52+ 122 = 25 + 144 = 169 = 132

b) The numbers 7, 24, 25

a2 + b2 = c2

72 + 242 = 25

because        72 = 7 x7 = 49

242 = 24 x 24 = 576

252 = 25 x 25 = 625

and so                72 + 242 = 49 + 576 = 625 = 25 2

The numbers above satisfy similar condition that  is pythagors theorem.  I have tested the numbers by putting them into the formula a2 + b2 = c2.  This is means the numbers are a Pythagorean triple because they satisfy the condition a2 + b 2 = c2

2)

 Length of shortest side Length of middle side Length of longest side Perimeter Area 3 4 5 12 6 5 12 13 30 30 7 24 25 56 84 9 40 41 90 180

a) I found the perimeter for the sequence 5, 12, 13 and 7, 24, 2, by simply finding the sum (add) all the lengths of the three sides together to get the perimeter.

E.g. Perimeter = shortest +middle + longest length

5 + 12 + 13 = 30 units

7 + 24 + 25 = 56 units

I found the area  for the sequences 5, 12, 13 and 7, 24, 25, by simply finding the product (multiply) the shortest and middle and halving the answer.

E.g. Area = ½ x shortest x middle length

½ x 5 x 12 = 30 square units

½ x 7x 24 = 84 square units

I used this method for the area because it is a right-angled triangle.  The dotted lines above show that it is actually half a square. Since the area of a square is length x width.  Therefor the area would be the same equation divided by 2.

Patterns

b)

Middle

if         1 x 4= 4

2 x 6 = 8

3 x 8 = 24

then        4 x 10 = 40

• From this information I could form a formula by finding the number that you multiply n with to get Un.  So I formed the formula 2n +n, that would equal the number that you multiply.  Once I found this I then  knew that by mutilplying it by n I could get Un.

E.g.         (2n +n) x n = Un

(2 x 2 + 2) x 2 = 12

(2 x 3 + 2) x 3 = 24

(2 x 4 + 2) x 4 = 40

Length of the Longest side

• The sequence was 5, 13, 25.  To find the next number in the sequence I used the same method as I did to find the middle length as they both followed similar patterns
 n Un 1st difference 2nd difference 1 5 8 2 13 4 12 3 25 4 16 4 41

The table shows that the first differences in the table again are not the same, therefor I then  worked out the second difference.  I found the second differences to be constant – all 4.  This pattern enabled me to calculate the fourth term (highlighted in blue).  I

added 4 to the first difference in series (4 + 12 = 16), the sum was then added to Un in series (16 + 24 = 40).

• Another pattern I noticed was that if you added 1 to the middle length of the same sequence then the longest length was obtained.

E.g.

if         4 + 1 = 5

12 + 1 = 13

24 + 1 = 25

then        40 + 1 = 41

I further checked that the 4th

Conclusion

tn  = 2n2  + 3                                                        tn  = 3n2  + 5n

There is a relationship between the second row of difference and a.  This is that the second row of difference halved is equal to a.  Therefor I predict that if the second row of difference is 4 then a will be equal to 2.

Finding the rule

I will investigate further to find b and c

I can then use this to find the quadratic rule for the sequence 4, 12, 24, 40….

I will investigate further to find ba and c.  For  tn  = an2 = bn + c.

4                12                24                40

8                12                16

1. 4

2a = 4                3a + b         = 8                a + b + c = 4

a   = 2                  3 x 2 + b = 8                2 + 2 + c = 4

6 + b         = 8                4 + c         = 4

b         = 8 – 6                c        = 4 – 4

b         = 2                 c        = 0

The rule is 2n2 + 2n

Testing and proving

• To test the formula is correct I will write down the first eight sequences given by the rule. I can then compare them with those I got using the patterns I noticed. Using tn = 2n2 + 2n.

t1 =        2 x 12 + 2 x 1                t2 =        2 x 22 + 2 x 2                t3 =        2 x 32 + 2 x 3

4                                12                                24

t4 =        2 x 42 + 2 x 4                t5=        2 x 52 + 2 x 5                t6=        2 x 62 + 2 x 6

40                                60                                84

• I will also test the formulae by using the rule to find the 8th  term of the sequence

Using tn = 2n2 + 2n, t8  =  2 x 82 + 2 x 8

= 144

•  I can also use the formula in algebra equation to find the term of the sequence 8.

GCSE MATHS COUSEWORK/BEYONG PYTHAGORS

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