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  • Level: GCSE
  • Subject: Maths
  • Document length: 1506 words

Patterns With Fractions Investigations

Extracts from this document...

Introduction

Mathematics Coursework. Patterns With Fractions. Consider the sequence of fractions and the differences between the fractions: Term (n) 1 2 3 4 5 1st Difference 2nd Difference (For rest of differences, see page11) Finding the starting fraction for the nth term: , , , , = (The general formula) Check if correct formula: Term (n) Numerator (n) Denominator (n + 1) Final Fraction 1 1 2 ? 2 2 3 ? 3 3 4 ? 4 4 5 ? 5 5 6 ? (Check On Page 11) Finding the nth term for the 1st difference: In order to find out the nth term for the 1st differences, the requirement is to subtract the 2nd fraction from the 1st fraction (the smaller fraction from the bigger fraction). - = = = (The general formula for 1st difference) Check if correct formula: Term (n) Numerator (1) Denominator (n + 1)(n+2) Final Fraction 1 1 (1+1)(1+2) = 6 ? 2 1 (2+1)(2+2) = 12 ? 3 1 (3+1)(3+2) = 20 ? 4 1 (4+1)(4+2) = 30 ? 5 1 (5+1)(5+2) = 42 ? (Check On Page 11) ...read more.

Middle

formula. Justification of the 2nd difference: The formula for the 2nd difference: In order to justify the formula, it is best to use the area of the formula that alters the most between formulas to formula. This is the case of the denominators in the formulas. The formula for the denominator is [(n+1)(n+2)(n+3)]. The actual denominators for the second differences are: Denominator 12 30 60 105 168 1st difference 18 30 45 63 2nd difference 12 15 18 3rd difference 3 3 As can be seen, in order to find a constant difference, the third difference for the denominators had to be found. This therefore shows that the general formula is: y = an3 + bn2 + cn + d In the above formula, there are four unknowns; these have to be worked out in order to justify the formula. Work out 1st unknown (the constant a) In order to work out the constant a, the rule for the 3rd difference must be used. a = of 3 a = Work out 2nd, 3rd and 4th (the constants b, c and d) Using the fact that a = , put the first constant in the formula y = an3 + bn2 + cn + d Formula No Term (n) ...read more.

Conclusion

= (n2 + 2n + n + 2)(n+3) = n3 + 2n2 + n2 + 2n + 3n2 + 6n + 3n + 6 = n3 + 6n2 + 11n + 6 Formula is correct (?) Finding the general term: So far, I have made formulas to interrogate the desired fraction through inputting the term number. A general term would be much easier to use if I could enter the term number and also the difference number and be left with the fraction that is required. Below are all of the different formulas for the different differences. > Starting Fraction > 1st Difference > 2nd Difference > 3rd Difference > 4th Difference > 5th Difference The general term for the fractions is: Check if correct formula: Term (n) Difference (x) Numerator (n! x!) Denominator (n + x + 1)! Fraction Final Fraction 6 4 720 * 24 (6 + 4 + 1)! 5 5 120 * 120 (5 + 5 + 1)! 4 2 24 * 2 (4 + 2 + 1)! 3 3 6 * 6 (3 + 3 + 1)! 2 1 2 * 1 (2 + 1 + 1)! (Check on page 11) Checking Sheet Starting Fraction 1st Difference 2nd Difference 3rd Difference 4th Difference 5th Difference Ahzaz Chowdhury 11H Fractions Coursework. - 1 - ...read more.

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