If f(x)=∑n=0∞x2n2n⋅n!f(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{2^n \cdot n!}f(x)=∑n=0∞2n⋅n!x2n, what is f′(x)f'(x)f′(x)?
∑n=1∞x2n−12n−1⋅(n−1)!\sum_{n=1}^{\infty} \frac{x^{2n-1}}{2^{n-1} \cdot (n-1)!}∑n=1∞2n−1⋅(n−1)!x2n−1
∑n=0∞x2n+12n⋅n!\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2^n \cdot n!}∑n=0∞2n⋅n!x2n+1
∑n=1∞n⋅x2n−12n⋅n!\sum_{n=1}^{\infty} \frac{n \cdot x^{2n-1}}{2^n \cdot n!}∑n=1∞2n⋅n!n⋅x2n−1
∑n=1∞2n⋅x2n−12n⋅n!\sum_{n=1}^{\infty} \frac{2n \cdot x^{2n-1}}{2^n \cdot n!}∑n=1∞2n⋅n!2n⋅x2n−1