Answer
$\sin \ \pi x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{\pi^{2n+1}(x)^{2n+1}}{(2n+1)!}$
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 $\pi x$ in the above series.
$\sin \pi x=\pi x-\dfrac{(\pi x)^3}{3!}+\dfrac{(\pi x)^5}{5!}+..... $
Thus, our required series is:
$\sin \ \pi x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{\pi^{2n+1}(x)^{2n+1}}{(2n+1)!}$
