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**PulpFiction1950s****Member**- Registered: 2006-11-06
- Posts: 4

I'm having problems with my nth terms and would like someone to help me with these questions. The nth terms I have to find are from fraction sequences shown below:

I have worked out the first sequences but if you notice any mistakes please point them out.

1/2 2/3 3/4 4/5 5/6 6/7 7/8 8/9 9/10 10/11 n/(n+1)

1/6 1/12 1/20 1/30 1/42 1/56 1/72 1/90 1/110 1/(n+1)(n+2)

1/12 1/30 1/60 1/105 1/168 1/252 1/360 1/495

1/20 1/60 1/140 1/280 1/504 1/840 1/1320

1/30 1/105 1/280 1/630 1/1260 1/2310

1/42 1/168 1/504 1/1260 1/2772

In addition to finding the nth term for each sequence, you can see that it makes a triangle shape and the entire thing is also one big sequence so I need to find the general nth term for the entire thing as well. If you could please walk me through this and help me find my way towards an answer I would be very grateful because this work is severly important. I know it may seem a lot but please get to me as soon as possible.

Thank you,

PulpFiction1950s

*Last edited by PulpFiction1950s (2006-11-07 07:04:18)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Sometimes you can find the nth term of a sequence by taking differences over and over again like that, but for that to happen the nth term needs to be a polynomial. If it's not, you'll just be taking differences for ever and never get anywhere.

It's fairly obvious by looking that the nth term of your sequence is just n/(n+1).

Why did the vector cross the road?

It wanted to be normal.

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**PulpFiction1950s****Member**- Registered: 2006-11-06
- Posts: 4

I have tried using the differences (doesn't work).

As for ure n/(n+1) I know this I want help with the other sequences and the nth term for the pyramid.

Thanks.

P.S. I may be sounding ungrateful (which I am not) but please hurry because my time is coming to an end with this work.

*Last edited by PulpFiction1950s (2006-11-07 07:05:27)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Oh, I see what you're doing now.

You can actually work out the nth term of each line of the pyramid fairly easily by thinking about what each of the terms actually is. The first term on the second line is the second term on the first line subtracted by the first term on the first line. More generally, the nth term on the second line is the (n+1)th term on the first line subtracted by the nth term on the 1st line.

More generally still, the nth term on the mth line is the (n+1)th term on the (m-1)th line subtracted by the nth term on the (m-1)th line.

You know the nth term of the 1st line, so you can use that to algebraically work out the nth term of the second line.

nth term {2}= (n+1)/(n+2) - n/n+1 = (n+1)² - n(n+2)/[(n+1)(n+2)] = 1/(n+1)(n+2), which is what you have.

Now you can use that to work out the nth term of the 3rd line.

nth term {3}= 1/(n+2)(n+3) - 1/(n+1)(n+2) = [(n+1)-(n+3)]/[(n+1)(n+2)(n+3)] = -2/(n+1)(n+2)(n+3). If you just want the difference and don't care about which of the two numbers on the row above is the bigger one, then you can get rid of the minus sign.

And then you can use that to work out the nth term of the 4th line, then use *that* to work out the nth term of the 5th line, etc.

Why did the vector cross the road?

It wanted to be normal.

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**PulpFiction1950s****Member**- Registered: 2006-11-06
- Posts: 4

Thank you very much.

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**PulpFiction1950s****Member**- Registered: 2006-11-06
- Posts: 4

Ok, I have one more problem that has arised. I have worked out the nth term forumla for each difference (thanks to your help). They are as follows:

1st difference:

1/(n+1)(n+2)

2nd difference:

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

3rd difference:

6/(n+1)(n+2)(n+3)(n+4)

and so on...

Now notice how the numerator goes like this:

1

then

2 (which is 1x2)

then

6 (which is 1x2x3)

My objective is to find the nth term for the entire pyramid of fractions (my first post) but I need help.

I know that it is something to do with the 1x2 and 1x2x3 and 1x2x3x4, etc but I don't know how to put this into a nth term forumla.

Please note, as always, I do not want a direct answer but instead I would just like to be pointed into the right direction so that I can find the answer for myself.

I have till tomorrow to finish this work so please send a reply today.

Thank you very much.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Do you know about factorial notation? That would help immensely here. It's a fairly well known sign, so you could easily find out about with Google or something.

Also, can the denominator stay as (n+1)(n+2)(n+3)...(n+(d+1)), where d is the line of difference, or do you want to make that neater as well? Factorial notation would probably help to neaten that up as well.

Why did the vector cross the road?

It wanted to be normal.

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

carrying on from mathsyperson, in factorial notation

1×2×3×...×y

-------------------------

(n+1)(n+2)...(n+y+1)

the numerator, can be written as y!

the denomiator, can be written as (n+y+1)! / n!

i.e.

if y = 2, you get

if y = 3, you get

The Beginning Of All Things To End.

The End Of All Things To Come.

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