• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
• Level: GCSE
• Subject: Maths
• Word count: 1506

# Patterns With Fractions Investigations

Extracts from this document...

Introduction

Ahzaz Chowdhury 11H

Fractions Coursework.

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) FinalFraction 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) FinalFraction 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)

Finding the nth term for the 2nd difference:

In order to find out the nth term for the 2nd differences, the requirement is to subtract the 1st fraction from the 2nd fraction (the smaller fraction from the bigger fraction) of the 1st differences.

-

=

=                (The general formula for 2nd difference)

Check if correct formula:

 Term(n) Numerator(2) Denominator(n + 1)(n+2)(n+3) Fraction FinalFraction 1 2 (1+1)(1+2)(1+3) =24 ✓ 2 2 (2+1)(2+2)(2+3) = 60 ✓ 3 2 (3+1)(3+2)(3+3) =120 ✓ 4 2 (4+1)(4+2)(4+3) =210 ✓ 5 2 (5+1)(5+2)(5+3) =504 ✓

(Check On Page 11)

Finding the nth term for the 3rd difference:

Middle

(Check On Page 11)

Finding the nth term for the 4th difference:

In order to find out the nth term for the 4th differences, the requirement is to subtract the 1st fraction from the 2nd fraction (the smaller fraction from the bigger fraction) of the 3rd differences.

-

=

=

=

(The general formula for 4th difference)

Check if correct formula:

 Term(n) Numerator(24) Denominator(n + 1)(n+2)(n+3)(n+4)(n+5) Fraction FinalFraction 1 24 (1+1)(1+2)(1+3)(1+4)(1+5) =720 ✓ 2 24 (2+1)(2+2)(2+3)(2+4)(2+5) = 2520 ✓ 3 24 (3+1)(3+2)(3+3)(3+4)(3+5) =6720 ✓ 4 24 (4+1)(4+2)(4+3)(4+4)(4+5) =15120 ✓ 5 24 (5+1)(5+2)(5+3)(5+4)(5+5) =30240 ✓

(Check On Page 11)

Finding the nth term for the 5th difference:

In order to find out the nth term for the 5th differences, the requirement is to subtract the 1st fraction from the 2nd fraction (the smaller fraction from the bigger fraction) of the 4th differences.

-

=

=

=

(The general formula for 5th difference)

Check if correct formula:

 Term(n) Numerator(120) Denominator(n+1)(n+2)(n+3)(n+4)(n+5)(n+6) Fraction FinalFraction 1 120 (1+1)(1+2)(1+3)(1+4)(1+5)(1+6) =5040 ✓ 2 120 (2+1)(2+2)(2+3)(2+4)(2+5)(2+6) = 20160 ✓ 3 120 (3+1)(3+2)(3+3)(3+4)(3+5)(3+6) =60480 ✓ 4 120 (4+1)(4+2)(4+3)(4+4)(4+5)(4+6) =151200 ✓ 5 120 (5+1)(5+2)(5+3)(5+4)(5+5)(5+6) =332640 ✓

(Check On Page 11)

In order to check if the formulas I have worked out are correct, it is best to justify one of the differences as justifying one of them proves that the rest of the formulas must be correct. In order make a fair generalisation; I am going to justify the 2nd difference (denominator) 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 formulay = an3 + bn2 + cn + d

 Formula No Term (n) y = an3 + bn2 + cn + d Final Formula 1 1 12 = 0.5 + b + c + d b + c + d  = 11.5 2 2 30 = 4 + 4b + 2c + d 4b + 2c + d  = 26 3 3 60 = 13.5 + 9b + 3c + d 9b + 3c + d  = 46.5

Conclusion

c5 c21">=      y = n3 + 3n2 + 5.5n + 3
 Term(n) y = n3 + 3n2 + 5.5n + 3 Denominator(y) 1 � + 3 + 5.5 + 3 12 ✓ 2 4 + 12 + 11 + 3 30 ✓ 3 13.5 + 27 + 16.5 + 3 60 ✓ 4 32 + 48 + 22 + 3 105 ✓ 5 62.5 + 75 + 27.5 + 3 168 ✓

* Note: The formula is correct. *

In order to transform y = n3 + 3n2 + 5.5n + 3 so that there is no longer any decimals or fractions in it, I shall multiply the denominator formula by 2.

2 [n3 + 3n2 + 5.5n + 3]

= n3 + 6n2 + 11n + 6

Check if above formula is correct:

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

= (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

-  -

This student written piece of work is one of many that can be found in our GCSE Hidden Faces and Cubes section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Hidden Faces and Cubes essays

1. ## I am doing an investigation to look at shapes made up of other shapes ...

the 'divide by' signs in the same place, even if they are only /1, as it may help me when trying to find a connection between the three. I have therefore rearranged the three formulas above to give: T=(P+2D-2)/1, Q=(P+2D-2)/2 and H=(P+2D-2)/4 Now that I have made the formulas look

2. ## Border coursework

N=3 N = {? 2(2n2 -2n+1)} + 2n2 -2n+1 The summation function ? for the range n=1 to n=n-1, works as follows: N={2+10} +13 N=25 Correct I will now use this formula to try for the next sequence N=4 Sequence 4 N = {?

1. ## shapes investigation coursework

the same (although underneath they still do exactly the same as before), it is obvious that the only difference between them is the amount that they are divided by (i.e. /1, /2 and /4). But why are these 'divided by' amounts different?

2. ## GCSE Maths Coursework - Shapes Investigation

For all three shapes, the values end in 2. This is -2 exists in the (P+2D-2) part of my universal formula. Once you have removed the extra 2 form each shape, you are left with a number which, when divided by the number of shapes (in this case 10)

1. ## Shapes Investigation I will try to find the relationship between the perimeter (in cm), ...

is out by -15 H=24+18-2 � H=40 DH is out by +30, or +3H, or x4H From these trials I can see that there is obviously a pattern with the H=P+2D-2 formula, in which the answer I get is always four times the answer I need.

2. ## gcse maths shapes investigation

subtracted in the formulas, a value of 0 will not alter the result without D included.

1. ## mathsI will try to find the correlations between the perimeter (in cm), dots enclosed ...

However, just to make sure, I will test the new formulas once for each number of triangles. I will make sure I do not test them with a D of zero, as this would give less margin for error (I have not tested any shapes where T or P are zero)

2. ## I am doing an investigation to look at shapes made up of other shapes.

P=11 and D=3 � 11-2+6=15 C P=13 and D=2 � 13-2+3=15 C P=15 and D=1 � 15-2+2=15 C P=17 and D=0 � 17-2+0=15 C And where T=16... P=10 and D=4 � 10-2+8=16 C P=12 and D=3 � 12-2+6=16 C P=14 and D=2 � 14-2+4=16 C P=16 and D=1 � 16-2+2=16

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to