Generating function of (-3)^n

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Given the sequence $a_n$, where $n$-th element is $a_n = (-3)^n$

I have the generating function $$A(t) = \sum_{n=0}^{\infty} (-3)^nt^n $$

The problem now is to simplify the obtained expression.

I did the following:

Let $B(t) = -3t$ be the generating function and $C(B(t)) = A(t) = \sum_{n=0}^{\infty} (-3)^nt^n $. Hence, the generating function $C(t) = \sum_{n=0}^{\infty} t^n = \frac{1}{1 - t}$

Therefore $$A(t) = C(B(t)) = C(-3t) = \frac{1}{1 + 3t}$$

Is this a possible way to solve this problem?

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You could simply notice that $\sum_{n=0}^\infty(-3)^nt^n$ is a convergent geometric series with ratio $-3t$ for $\|3t\|< 1$.

The closed form of geometric series with ratio $r$, where $\|r\| < 1$, is $$ \sum_{n=0}^\infty r^n = \frac{1}{1-r}. $$