Answer
$\sin \ 3x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(3x)^{2n+1}}{(2n+1)!}$
Work Step by Step
We know that the Maclaurin series for $\sin x=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+..... $
Replace $x$ with $3x$ in the above series.
$\sin 3x=3x-\dfrac{(3x)^3}{3!}+\dfrac{(3x)^5}{5!}+..... $
Thus, our required series is:
$\sin \ 3x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(3x)^{2n+1}}{(2n+1)!}$
