Answer
Centred at $0$.
Work Step by Step
Re-write the given series in the expanded form as follows:
$\Sigma_{n=1}^\infty \dfrac{(-1)^n 1 \cdot 3 \cdot 3 \cdot ........(2n-1) x^n}{2^n n!}=\dfrac{-(x-0)}{2}+\dfrac{3(x-0)^2}{2^2 \cdot 2!}+\dfrac{-5(x-0)^3}{2^3 \cdot 3!}+.....$
We can see that the above power series is centred at $0$.
