This is a geometric sequence because it has a common ratio, denoted r, meaning that each term is a multiple of the previous term.
[tex]a(n)=ar^{n-1}[/tex], a=initial term, r=common ratio, n=term number
In this case a=[tex]1 \over 81[/tex] and r=3 (because each term is 3 time the previous term) so
[tex]a(n)=(1/81)*3^{n-1}[/tex]
So the 14th term is:
a(14)=(1/81)[tex]*3^{14}[/tex]=59049
