Answer
$ \sin 3x=\Sigma_{n=0}^\infty \dfrac{(-1)^n 3^{2n+1}x^{2n+1}}{(2n+1)!}$ or, $ \sin 3x=3x-\dfrac{3^3x^3}{3!}+\dfrac{3^5x^5}{5!}$
Work Step by Step
Since, we know that the Maclaurin Series for $\sin{x}$ is defined as:
$ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$
We need to replace in the above series $x$ with $3x$.
we have
$ \sin 3x=\Sigma_{n=0}^\infty \dfrac{(-1)^n 3^{2n+1}x^{2n+1}}{(2n+1)!}$ or, $ \sin 3x=3x-\dfrac{3^3x^3}{3!}+\dfrac{3^5x^5}{5!}$
