# Maths algerbra

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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 X3

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.

Middle

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)

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

X (n + 3) (n+1)(n+2) (n+2)(n+3) X (n + 1)

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

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

X | n | 3 |

1 | n | 3 |

X | n | 1 |

1 | n | 1 |

(n + 3) – (n + 1)

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

which cancelled to

(n + 3) - (n + 1)

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

2

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

I tested this by using n = 3

2

(3 + 1)(3+2)(3+3)

2

4 X 5 X 6

2

120

1

60

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

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

I had to make the denominators the same so I …

Conclusion

X1

X2

X3

X4

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

?? ?? 12 24

2 X 3 X 4 X 5 X 6 3360 3360 720

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

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

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