Since $\displaystyle \sin{x} = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$

Then $\displaystyle 3\sin^2\dfrac{x}{2} = 3\bigg(\dfrac{x}{2}-\dfrac{\left(\dfrac{x}{2}\right)^3}{3!}+\dfrac{\le ft(\dfrac{x}{2}\right)^5}{5!}-\dfrac{\left(\dfrac{x}{2}\right)^7}{7!}+...\bigg) = 3\bigg(\dfrac{x}{2}-\dfrac{\dfrac{x}{2^3}^3}{3!}+\dfrac{\dfrac{x}{2^5} ^5}{5!}-\dfrac{\dfrac{x}{2^7}^7}{7!}+...\bigg) $

$\displaystyle = 3\bigg(\dfrac{x}{2}-\dfrac{x^3}{{2^3}3!}+\dfrac{x^5}{2^55!}-\dfrac{{x}^7}{{2}^77!}+...\bigg) = \dfrac{3x}{2}-\dfrac{3x^3}{{2^3}3!}+\dfrac{3x^5}{2^55!}-\dfrac{{3x}^7}{{2}^77!}+...+ \dfrac{(-1)^{n}(3x^{2n+1})}{2^{2n+1}(2n+1)!}-...$

$\displaystyle \Rightarrow$ $\displaystyle 3\sin^2{\dfrac{x}{2}} = \sum_{n=0}^{\infty}\dfrac{(-1)^{n}(3x^{2n+1})}{2^{2n+1}(2n+1)!}$.