Melody,

First I want to say thank you very much, you have helped me see the numerators as powers of 3.

I did some aditional study on the net, and came to this conclusion:

The numerators change to:

\(3^0;3^1;3^2;3^3\)

so this has to be :

\(3^{n-1}\) = NOMINATOR

which I understand...THEN:

you subtracted the denominators and found:

\(5;8;13;20\)

becomes: \(3;5;7;9\)

which becomes: \(2;2;2\)

From this we see that: 2a=2, therefore, a=1

3a+b = 3, therefore "b" calculates to 0

and a+b+c=T1=5, so "c" calculates to 4

The general equation is: \(an^2+bn+c\)

So substitute everything above into this equation, and it calculates to \(n^2+4\) = DENOMINATOR

So that is then \({Nominator \over Denominator}={3^{n-1} \over n^2+4}\)

thank you kindly for your help Melody!!