Example 1. A function of the natural number $\mathrm{n}$ satisfies the following conditions:
(1) $f(n)=f(n-1)+a^{n}(n \geqslant 2$, $a$ is a non-zero constant $)$;
(2) $f(1)=18$.

Try to find: (1) the analytical expression of $\mathrm{f}(\mathrm{n})$;
(2) when $a=1$, what is the value of $f(1983)$?
From the given recurrence relation, we get
$$
\begin{array}{l}
f(2)=f(1)+a^{2}, \\
f(3)=f(2)+a^{3}, \\
f(4)=f(3)+a^{4}, \\
\cdots \cdots \\
f(n-1)=f(n-2)+a^{n-1}, \\
f(n)=f(n-1)+a^{n} .
\end{array}
$$

By adding both sides of the above equations, we get
$$
\begin{array}{l}
f(n)=f(1)+a^{2}+a^{3}+a^{4}+\cdots+a^{n}, \text { thus } \\
f(n)=\left\{\begin{array}{l}
f(1)+\frac{a^{2}\left(a^{n-1}-1\right)}{a-1},(a \neq 1) \\
f(1)+(n-1) \cdot(a=1)
\end{array}\right.
\end{array}
$$
(2) $\mathrm{f}(1983)=18+(1983-1)=2000$.