Answer
$\cos (x^{3/2}) =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{3n}}{(2n)!}$
Work Step by Step
We know that the Maclaurin series for $\cos x=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+..... $
Replace $x$ with $x^{3/2}$ in the above series.
$\cos (x^{3/2})=1-\dfrac{(x^{3/2})^2}{2!}+\dfrac{(x^{3/2})^4}{4!}-\dfrac{(x^{3/2})^6}{6!}+..... =1-\dfrac{x^3}{2}+\dfrac{ x^6}{24}-.......$
Thus, our required series is:
$\cos (x^{3/2}) =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{3n}}{(2n)!}$
