No Guest, you have not interpreted Juriemagic's question properly.
He has the answer but he wants to know how he could have seen it, or worked it out by himself.
Lets look at the numerators seperately from the denominators
numerator - I notice straight off that these are powers of 3
n | 1 | 2 | 3 | 4 | n |
T_n | 1 | 3 | 9 | 27 | |
3^(1-1) | 3^(2-1) | 3^(3-1) | 3^(4-1) | 3^(n-1) |
denominator
5 8 13 20
now I am going to subtract the preceding term from each of these.
8-5=3 13-8=5 20-13=7
3 5 7 If these had all been the same then it could have been expresed as polynomial of degree 1
Now I wll subtract again
5-3=2 7-5=2
2 2 These are the same so I can express this sequence as a polynomial of degree 2
n | 1 | 2 | 3 | 4 | n |
n^2 | 1 | 4 | 9 | 16 | n^2 |
t_n | 5 | 8 | 13 | 20 | |
t^n+4 | 5 | 8 | 13 | 20 |
So putting this altogether I have found the sequence to be.
\(T_n=\dfrac{3^{n-1}}{n^2+4}\)
If you have any questions Juriemagic just ask. If you want to be sure that I see it, send me a personal message and include the address of the question. :)