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It looks like the series could be
[tex]\displaystyle 1 + 3x + \frac{9x^2}{2!} + \frac{27x^3}{3!} + \cdots = \sum_{n=0}^\infty \frac{3^nx^n}{n!} = \sum_{n=0}^\infty \frac{(3x)^n}{n!}[/tex]
Recall that
[tex]\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}[/tex]
It follows that the given series is the power series expansion for [tex]\boxed{e^{3x}}[/tex].