Coursework regarding fraction differences (sequence with nth term- fractions)

**math-** · December 12th 2007, 11:45 AM

hi everyone!

ok, i have an investigation where i must consider some fraction sequences.

first row- 1/2, 2/3, 3/4, 4/5 5/6, 6/7, 78

second row- 1/6, 1/12, 1/20, 1/30, 1/42, 1/56

third row- 1/12, 1/30, 1/60, 1/105, 1/168

fourth- 1/20, 1/60, 1/140, 1/280

(there are 3 more rows aswell)

my task is to find out the nth term for each row(rule) and explain why the rule works algebraically.
the relationship between the rows is the first two between the fractions are that the two fractions above are minused which gives you that fraction.(it becomes obvious when you look.)

i have figured out the rule for the first row--- n/(n+1)

i just looked and could see that the numerator was exactly the same as n, and the denominator was n+1.

and thats all i have done.
i dont know how to figure out the rest, if anyone out there knows then that would be very much appriciated!
cheers

if you need some more description just say

**Geometor** · December 12th 2007, 01:24 PM

Well as a tip
i think it it best to just focus on the denominator because by the looks of it all of them will be 1/ something. So let's make it:

second row- 6, 12, 20, 30, 42, 56

third row- 12, 30, 60, 105, 168

fourth- 20, 60, 140, 280

find the nth term for these and you're done

Sorry I haven't come up with any answers atm

**earboth** · December 12th 2007, 10:42 PM

Originally Posted by math-

hi everyone!

ok, i have an investigation where i must consider some fraction sequences.

first row- 1/2, 2/3, 3/4, 4/5 5/6, 6/7, 78

second row- 1/6, 1/12, 1/20, 1/30, 1/42, 1/56

third row- 1/12, 1/30, 1/60, 1/105, 1/168

fourth- 1/20, 1/60, 1/140, 1/280

...

Hello,

as Geometor suggested I'll use the denominators only:

Consider the following sequence and calculate the diffferences between the numbers:

Code:

 6   12   20   30   42   56
    6   8   10   12   14
      2   2    2    2

The differences of the differences yield a constant sequence. Thus your sequence is of the second grade/degree(?) and the general equation of such a sequence is:

$s(n)=a \cdot n^2 + b \cdot n + c$
You know

$ \begin{array}{l}s(1) = 6\\s(2)=12\\s(3)=20\end{array}~\implies~ \left|\begin{array}{lcr}a+b+c&=&6\\ 4a+2b+c&=&12\\9a+3b+c&=&20\end{array}\right.$

Solve this system of simultaneous equations. I've got: a = 1, b = 3, c = 2

That means the sequence have the equation

$s(n) = n^2 + 3n + 2$

And the numbers in the second row can be calculated by:

$r_2(n) = \frac1{n^2 + 3n + 2}$

If you apply this method to your 3rd row you'll get a sequence of third grade/degree(?). Thus you have to solve a system of 4 simultaneous equations:

$ \begin{array}{l}s(1) = 12\\s(2)=30\\s(3)=60\\s(4) = 105\end{array} ~\implies~ \left|\begin{array}{lcr}a+b+c+d&=&12\\ 8a+4b+2c+d&=&30\\27a+9b+3c+d&=&60\\ 64a+16b+4c+d&=&105\end{array}\right.$

Solve this system. I've got as the final result that the numbers in the third row can be calculated by:

$r_3(n) = \frac1{\frac12 \cdot n^3 + 3n^2 + \frac{11}2 \cdot n +3}$

**earboth** · December 13th 2007, 01:29 AM

Originally Posted by math-

hi everyone!

ok, i have an investigation where i must consider some fraction sequences.

first row- 1/2, 2/3, 3/4, 4/5 5/6, 6/7, 78

second row- 1/6, 1/12, 1/20, 1/30, 1/42, 1/56

...

Hello,

sometimes it takes some time ...

The numbers in the 2nd row are the differences of the numbers in the 1rst row. Using your result:

$\frac{n+1}{n+1+1}-\frac{n}{n+1}= \frac{(n+1)^2-n(n+2)}{(n+2)(n+1)} = \frac{n^2+2n+1-n^2-2n}{n^2+3n+2}=\frac1{n^2+3n+2}$

It's the same result I published in my previous post but the way to get it is a little bit more Soroban-like

For the 3rd row try:

$\frac1{(n+2)(n+1)}-\frac1{(n+3)(n+2)}= \frac{n+3-n-1}{(n+3)(n+2)(n+1)}= \frac{2}{(n+3)(n+2)(n+1)}$

Go on and try if this method will do for the next rows. Good luck!

**Soroban** · December 13th 2007, 10:03 AM

Hello, math-!

I have an investigation where I must consider some fraction sequences.

$\begin{array}{cccccccc} 1^{st}\text{ row:} & \frac{1}{2} & & \frac{2}{3} & & \frac{3}{4} & & \frac{4}{5} \\ 2^{nd}\text{ row:} & & \frac{1}{6} & & \frac{1}{12} & & \frac{1}{20} \\ 3^{rd}\text{ row:} & & & \frac{1}{12} & & \frac{1}{30} & & \frac{1}{60} \\ 4^{th}\text{ row:} & & & & \frac{1}{20} & & \frac{1}{60} \end{array}$ $\begin{array}{cccccc} & \frac{5}{6} & & \frac{6}{7} & & \frac{7}{8} \\ \frac{1}{30} & & \frac{1}{42} & & \frac{1}{56} \\ & \frac{1}{105} & & \frac{1}{168} \\ \frac{1}{140} & & \frac{1}{280} \end{array}$

(There are 3 more rows as well.)

You are correct: the general term of the first row is: . $f_1(n) \:=\:\frac{n}{n+1}$

In the second row, the general term is the difference of the two fractions directly above it.

. . Hence: . $f_2(n) \;=\;\frac{n+1}{n+2} - \frac{n}{n+1} \:=\:\frac{1}{(n+1)(n+2)}$

In the third row, the general term is the difference of the two fractions directly above it.

. . Hence: . $f_3(n) \;=\;\frac{1}{(n+1)(n+2)} - \frac{1}{(n+2)(n+3)} \;=\;\frac{2}{(n+1)(n+2)(n+3)}$

And so on . . .

**math-** · December 13th 2007, 10:06 AM

Originally Posted by Soroban

Hello, math-!

You are correct: the general term of the first row is: . $f_1(n) \:=\:\frac{n}{n+1}$

In the second row, the general term is the difference of the two fractions directly above it.

. . Hence: . $f_2(n) \;=\;\frac{n+1}{n+2} - \frac{n}{n+1} \:=\:\frac{1}{(n+1)(n+2)}$

In the third row, the general term is the difference of the two fractions directly above it.

. . Hence: . $f_3(n) \;=\;\frac{1}{(n+1)(n+2)} - \frac{1}{(n+2)(n+3)} \;=\;\frac{2}{(n+1)(n+2)(n+3)}$

And so on . . .

Hi
thanks for all the replies.
i have already worked out the second to being 1/(n+2)(n+3)
but i do not really no how to work out the third row which means i dont really understand how i could apply this to the 4rth row. if you could point me in the right direction(i found out the difference of the fractions to get the rule)
cheers

**math-** · December 18th 2007, 08:20 AM

thanks for all your help so far!!!!

i have succesfully worked out the differences for the first 6 rows.
however, i have to explain algebriacally, in other words justify why the forumla works for the second row with the forumla 1/(n+1)(n+2)(n+3).
i am only going to look at the demominator as this is the main thing which is changing throughout the rules.
but that is as far as i am getting an i do not know how to progress from here.

any help would great!!
and again thanks for all the help so far.
cheers

dont worry about that, i have already figured it out.

but i need to gain a general formula for the formulas relating to factorials.

cheers

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