Answer
$-5x+\dfrac{5}{3!}x^3-\dfrac{5}{5!}x^5+\dfrac{5}{7!}x^7+...$
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)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Now, $ 5\sin(- x)=5[(-x)-\dfrac{(-x)^3}{3!}+\dfrac{(-x)^5}{5!}-\dfrac{(-x)^7}{7!}+...]$
or, $=-5x+\dfrac{5}{3!}x^3-\dfrac{5}{5!}x^5+\dfrac{5}{7!}x^7+...$
