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  • Level: GCSE
  • Subject: Maths
  • Word count: 1208

Maths algerbra

Extracts from this document...

Introduction

Maths coursework

In  this coursework we were asked to find the differences in a simple fraction pattern. We had ;

Number

1

2

3

4

5

6

7

Fraction

1/2

2/3

3/4

4/5

5/6

6/7

7/8

I will look at the differences between the fractions and search  for any general rules. I will apply my good algebraic skills to help find any rules.

I then had to find the differences by taking the fractions away from each other. To find my first difference I had to ;

D1 = difference one

Fraction         2nd-1st                 3rd-2nd                4th-3rd

D1                  1st                               2nd                              3rd

I had to take away 2/3 form  1/2 but the  denominators were not the same, so I found the lowest common multiple  from each and multiplied them top and bottom eg.

X2   2   -     1   X3image01.pngimage00.pngimage01.png

X2   3         2   X3

        The lowest common  multiple is 6 so I multiplied  2/3 by 2 and 1/2 by 3. I then got ;

4/6 -  3/6 which then equalled 1/6, which was my 1st difference.

I done the same to 3/4 -  2/3             9/12 – 8/12 which equalled 1/12 and my second difference.image14.png

...read more.

Middle

nd – 3rd                 3rd   – 4th

D2           1st                               2nd                               3rd

I then looked for the nth term in the second difference, I had to take the nth term away from the n + 1th  term and so I …

            1          -           1                

  (n+1)(n+2)       (n+2)(n+3)   image02.pngimage02.png

X (n + 3)             1      -           1                 X  (n + 1)image15.pngimage02.png

X (n + 3)  (n+1)(n+2)       (n+2)(n+3)   X  (n + 1)  image09.png

            1X (n+3)                         -    1 X  (n + 1)

image16.pngimage17.png

(n + 1)(n+2)(n+3)       (n + 1)(n+2)(n+3)   image09.png

X

n

3

1

n

3

X

n

1

1

n

1


image18.png

        (n + 3) – (n + 1)image19.png

        (n + 1) (n+2)(n+3)       image20.png

        which cancelled to

image07.pngimage07.png

                                        (n + 3)  - (n + 1)image22.pngimage22.png

image23.png

        (n + 1)(n+2)(n+3)       image24.png

        2

image16.png

                   (n + 1)(n+2)(n+3)

I tested this by using n = 3

                                    2

image16.png

                   (3 + 1)(3+2)(3+3)

image25.png

                                2

image26.png

                                                       4 X 5 X 6

image20.png

        2

image26.png

                                                                                     120image25.png

image28.png

        1image26.png

                                                                                                         60image29.png

This is the 3rd difference in the second difference line.

I went on to find the 3rd difference.

Number

1

2

3

4

5

n

n+1

D2

1/12

1/30

1/60

1/105

2/(n+1)(n+2)(n+3)

2/(n+2)(n+3)(n+4)

D3

1/20

1/60

1/60

1/140

I found this by taking the nth term away from the n+1th term ;

             2                                             2        image30.pngimage30.pngimage30.png

(n+1)(n+2)(n+3)         -       (n+2)(n+3)(n+4)

I had to make the denominators the same so I …

...read more.

Conclusion

X1

X2

X3image07.png

                X4
image40.png

I predicted that the numorator would be 4 times more then the previous numerator because of the patten forming (24), I also predicted the denominator be times by one more then the number difference. To check this I had to test it I put n = 1

image07.png

         ??        ??        12        24image41.png

2 X 3 X 4 X 5 X 6                         3360        3360        720image42.pngimage43.pngimage42.pngimage43.pngimage43.pngimage03.png

        1

        30

                                                                     this 1makes my prediction right

I went on to find dk I predicted it to be

   1 X 2 X 3 X 4… X k

(n+1)(n+2)(n+3)(n+4)…(n+1+k)

I proved this by ;

Numerator

Difference

D  1

D  2

D  3

D  4

D  k

fraction

1

1X  2

1X2X 3

1X2X3X 4

1X2X3X 4…X k

numerator

1

2

6

24

1X2X3X4..Xk

The difference number was the last consecutive number multiplied to get the numerator

Denominator

Difference

D  1

D  2

D  3

D  k

Actual fraction

(n+1)(n+2)

(n+1)(n+2)

(n+3)

(n+1)(n+2)

(n+3)(n+4)

(n+1)(n+2)

(n+3)(n+4)

…(n+K+1)

explanation

(n+1)

(n+1+1)

(n+1)(n+2)

(n+2+1)

(n+1)(n+2)

(n+3)

(n+3+1)

(n+1)(n+2)

(n+3)(n+4)

(n+k+1)

This is how I explain the denominator of Dk.

        Daisy Driscoll

                   10BA

...read more.

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