Courseworks

Fraction Differences

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 as the denominators corresponded to the implied first line as shown in this table below:nth number12345678Denominators12345678I shall call this Formula 1 (F1) for easy reference.Second SequenceAgain 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 and , and all the way up toI set the differences out in a table to try to find the pattern:nth number123456789Sequence2612203042567290First Difference4681012141618Second Difference22222222As there was a constant difference of 2 I believed that the formula would include n�. I applied this to the first number in the sequence ‘2’. So n� = (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 + 1I 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 n� I looked at the difference between n� and the numbers in the sequence:Sequencen�Difference21 x 1+ 162 x 2+ 2123 x 3+ 3204 x 4+ 4305 x 5+ 5By looking at this I saw that the formula was n� + n.The formula n� + 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 n� 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) = 2Then 2:2(2 + 1) = 6As these had worked I entered the rest in a table:n123456n(n + 1)2612203042The formula n(n + 1) was successful in predicting the next numbers in the sequence. This is Formula 3 (F3).Third SequenceFor 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.As with the second sequence I used a table to investigate the differences to see if I could see a pattern:nth number12345678Sequence3123060105168252360First Difference91830456384108Second Difference91215182124Third Difference33333Due to the constant appearance of 3 in this sequence, in the third difference, I knew there would be an n� 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 n�:Sequencen�Difference31+ 2128+ 43027+ 36064+ 4105125+ 20There 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 n� 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 = 6This 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:First attempt, I tried substituting n with 2:The answer was the next number in the sequence, I continued this time replacing n with 3:The formula had worked again, to make absolutely sure I decided to try the formula out two more times:With n as 4:With n as 5:I decided that this was enough to state that the formula was successful, and I call this formula (F4).Fourth SequenceAs 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: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) = 24The 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 numberI 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:Substituting n for 3:Substituting n for 4: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:FactorialsI 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 F3n(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 22 x 33 x 4Only it is going up by one each time.I looked at factorials with F4n(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 32 x 3 x 43 x 4 x 5This 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 = = 0F3 = = 1 (1)F4 = = 2 (1 x 2)F5 = = 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:And then worked my formula upon it, using the next factorial 24 (1 x 2 x 3 x 4) as the denominator.F6 == 5= 30It worked.Beginning AgainNow 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 number12345678Sequence246810121416First Difference2222222As 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 number12345678Sequence3691215182124First Difference3333333I saw that the formula would be 3n, 3 the constant first difference would be the coefficient of n.Another pattern had been found.In conclusion I would say that sequences can often throw up many surprises.

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