# Fraction Differences

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Fraction Differences

First Sequence

To begin with I looked at the first sequence of fractions to discover
the formula that explained it. As all the numerators were 1 I looked
at the denominators. As these all increased by 1 every time, I
figured that the formula was simply [IMAGE] as the denominators
corresponded to the implied first line as shown in this table below:

nth number

1

2

3

4

5

6

7

8

Denominators

1

2

3

4

5

6

7

8

I shall call this Formula 1 (F1) for easy reference.

Second Sequence

[IMAGE] [IMAGE] [IMAGE] [IMAGE] [IMAGE]

Again I decided to discount the numerator as it was 1, and I decided
to concentrate on the differences between the denominators rather than
the ‘fractions’. So I am looking for a formula that will explain the
sequence: 2, 6, 12, 20, 30.

First of all though I decided to extend the sequence in order to have
a broader range to work with. I used a calculator to work out the
following denominators finding the difference between [IMAGE] and
[IMAGE], [IMAGE] and [IMAGE] all the way up to [IMAGE]

I set the differences out in a table to try to find the pattern:

nth number

1

2

3

4

5

6

7

8

9

Sequence

2

6

12

20

30

42

56

72

90

First Difference

4

6

8

10

12

14

16

18

Second Difference

2

2

2

2

2

2

2

2

MLA Citation:
"Fraction Differences." 123HelpMe.com. 07 Apr 2020
<https://www.123helpme.com/view.asp?id=148360>.

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### Popular Essays

As there was a constant difference of 2 I believed that the formula
would include n². I applied this to the first number in the sequence
‘2’. So n² = (1 x 1 = 1). To get the first number of the sequence –
2 I would have to add 1. Therefore the formula could be: n2 + 1

I tried this formula for the second number in the series. 22 = 2 x 2 =
4 + 1 = 5.

My formula was wrong.

As I knew that the formula would include n² I looked at the difference
between n² and the numbers in the sequence:

Sequence

Difference

2

1 x 1

+ 1

6

2 x 2

+ 2

12

3 x 3

+ 3

20

4 x 4

+ 4

30

5 x 5

+ 5

By looking at this I saw that the formula was n² + n.

The formula n² + n worked. It successfully was able to predict the
next numbers in the sequence, this I call Formula 2 (F2).

Knowing that this formula worked, I decided to try altering the
formula in a way where the basic elements were kept but laid out in a
different way, I wondered if there would be another formula that
worked.

As n² with a co-efficient (a number or symbol multiplied with a
variable) could be written with brackets as n(n + 1) I decided to try
and see if this would produce the same results:

I began at the first step, replacing n with 1.

1(1 + 1) = 2

Then 2:

2(2 + 1) = 6

As these had worked I entered the rest in a table:

n

1

2

3

4

5

6

n(n + 1)

2

6

12

20

30

42

The formula n(n + 1) was successful in predicting the next numbers in
the sequence. This is Formula 3 (F3).

Third Sequence

For the third sequence we had only been given two fractions, so first
of all I decided to extend this sequence in order to illuminate any
patterns that may occur there. I did this by looking at the
differences between the previous fractions.

[IMAGE] [IMAGE] [IMAGE] [IMAGE] [IMAGE] [IMAGE] [IMAGE] [IMAGE]

As with the second sequence I used a table to investigate the
differences to see if I could see a pattern:

nth number

1

2

3

4

5

6

7

8

Sequence

3

12

30

60

105

168

252

360

First Difference

9

18

30

45

63

84

108

Second Difference

9

12

15

18

21

24

Third Difference

3

3

3

3

3

Due to the constant appearance of 3 in this sequence, in the third
difference, I knew there would be an n³ in the formula.

I also believed that the formula would be in some way connected with
one of the previous formulas and decided to test this by working from
the same formula as the previous one:

Again I looked at the difference between the numbers in the sequence
and n³:

Sequence

Difference

3

1

+ 2

12

8

+ 4

30

27

+ 3

60

64

+ 4

105

125

+ 20

There was no decisive pattern, the coefficient was neither a constant
number or an ‘n’ term.

I decided to look at the second correct formula I had deduced for the
second sequence n(n + 1) (F2) and thought about how I could extend
this to include the n³ I knew was necessary for the formula.

n(n + 1) (n + 2)

I started with the number 1:

1(1 + 1) (1 + 2) = 1(2)(3) = 1 x 2 x 3 = 6

This was twice what I needed (3 for the first number in the sequence)
so I divided it by 2.

So the formula I was going to attempt was:

[IMAGE]

First attempt, I tried substituting n with 2:

[IMAGE]

The answer was the next number in the sequence, I continued this time
replacing n with 3:

[IMAGE]

The formula had worked again, to make absolutely sure I decided to try
the formula out two more times:

With n as 4:

[IMAGE]

With n as 5:

[IMAGE]

I decided that this was enough to state that the formula [IMAGE] was
successful, and I call this formula (F4).

Fourth Sequence

As I had hit on a vague pattern with the formulas I decided to
investigate further by working out the differences between the last
sequence to produce a new one:

[IMAGE] [IMAGE] [IMAGE] [IMAGE] [IMAGE] [IMAGE] [IMAGE]

And then I applied the same formula pattern as before:

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

And substituted the n for 1:

1(1 + 1)(1 + 2)(1 + 3) = 24

The answer I had was 24. The sequence that I was working from gave me
4 as the first number, this would imply that I would have to divide by
6 to get my first number [IMAGE]

I decided to test this formula substituting the n for 2, 3 and 4.
Because I had been having success with this formula I decided that the
amount I would have to test could be lowered.

Substituting n for 2:

[IMAGE]

Substituting n for 3:

[IMAGE]

Substituting n for 4:

[IMAGE]

All three answers were correct, corresponding to the next three
numbers in the sequence. I state that this formula is successful and
I am calling it F5: [IMAGE]

Factorials

I saw that the definition of a factorial could be relevant to my
formulas that I had discovered so far. As a factorial is “the product
of all the positive integers from 1 to a given number” Online
Dictionary.

I saw that my formula was doing something similar to this, only going
up by one each time.

If we take F3

n(n + 1)

replace with the numbers:

1(1 +1)

2(2+1)

3(3+1)

We can see that this is actually factorials:

1 x 2

2 x 3

3 x 4

Only it is going up by one each time.

I looked at factorials with F4

n(n + 1)(n + 2)

replaced with numbers:

1(1+1)(1+2)

2(2+1)(2+2)

3(3+1)(3+2)

Simplified:

1 x 2 x 3

2 x 3 x 4

3 x 4 x 5

This was the same.

This wasn’t the only area where I had noticed factorials being used
though. By looking at my formulas and the denominators that were
implied here I saw that they were the first three factorials:

F1 = [IMAGE] = 0

F3 = [IMAGE] = 1 (1)

F4 = [IMAGE] = 2 (1 x 2)

F5 = [IMAGE] = 6 (1 x 2 x 3)

I decided to continue finding the difference between the previous
sequence to see if it would produce the next factorial – 24:

First I created the sequence by again finding the differences between
the previous fractions:

[IMAGE] [IMAGE] [IMAGE] [IMAGE]

And then worked my formula upon it, using the next factorial 24 (1 x 2
x 3 x 4) as the denominator.

F6 = [IMAGE]

[IMAGE] = 5

[IMAGE] = 30

It worked.

Beginning Again

Now that I had worked out formulas and found the pattern that they too
were ‘sequencing’ by I decided to look at a different set of fractions
to see if the patterns applied to them too. Whereas the first
sequence had the denominators: 1, 2, 3, 4, 5… I decided to try
denominators that went up by 2: 2, 4, 6, 8, 10…

I began by setting out the numbers in a table to find the differences:

nth number

1

2

3

4

5

6

7

8

Sequence

2

4

6

8

10

12

14

16

First Difference

2

2

2

2

2

2

2

As the first difference was a constant I knew that the formula would
contain this difference as the coefficient of n: 2n.

I now tried a sequence where the numbers went up by 3:

nth number

1

2

3

4

5

6

7

8

Sequence

3

6

9

12

15

18

21

24

First Difference

3

3

3

3

3

3

3

I saw that the formula would be 3n, 3 the constant first difference
would be the coefficient of n.