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).

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.

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.

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.