Given the function f(x)=11−x=∑n=0∞xnf(x) = \frac{1}{1-x} = \sum_{n=0}^{\infty} x^nf(x)=1−x1=∑n=0∞xn, what is the series representation of g(x)=11−2xg(x) = \frac{1}{1-2x}g(x)=1−2x1?
∑n=0∞(2x)n\sum_{n=0}^{\infty} (2x)^n∑n=0∞(2x)n
∑n=0∞2xn\sum_{n=0}^{\infty} 2x^n∑n=0∞2xn
∑n=0∞x2n\sum_{n=0}^{\infty} x^{2n}∑n=0∞x2n
∑n=0∞xn2n\sum_{n=0}^{\infty} \frac{x^n}{2^n}∑n=0∞2nxn