The functional form the given equation can be written as
\(\displaystyle{f{{\left({t}\right)}}}={1800}{\left({1.11}\right)}^{{t}}\)
where f(0)=1800 and 1.11 is the ratio of successive values of the function.

That is, \(\displaystyle\frac{{f{{\left({1}\right)}}}}{{f{{\left({0}\right)}}}}=\frac{{f{{\left({2}\right)}}}}{{f{{\left({1}\right)}}}}=\ldots=\frac{{{f{{\left({t}+{1}\right)}}}}}{{f{{\left({t}\right)}}}}=\ldots={1.11}\)

So the recursive rule for this function van be written as \(\displaystyle{f{{\left({t}+{1}\right)}}}={1.11}\cdot{f{{\left({t}\right)}}}\)

That is, \(\displaystyle\frac{{f{{\left({1}\right)}}}}{{f{{\left({0}\right)}}}}=\frac{{f{{\left({2}\right)}}}}{{f{{\left({1}\right)}}}}=\ldots=\frac{{{f{{\left({t}+{1}\right)}}}}}{{f{{\left({t}\right)}}}}=\ldots={1.11}\)

So the recursive rule for this function van be written as \(\displaystyle{f{{\left({t}+{1}\right)}}}={1.11}\cdot{f{{\left({t}\right)}}}\)