Find the Taylor series and the interval of convergence for f(x)=1x+2f(x) = \frac{1}{x+2}f(x)=x+21 centered at a=2a = 2a=2.
∑n=0∞(−1)n(x−2)n4n+1\sum_{n=0}^{\infty} \frac{(-1)^n (x-2)^n}{4^{n+1}}∑n=0∞4n+1(−1)n(x−2)n with interval (−2,6)(-2, 6)(−2,6)
∑n=0∞(−1)n(x−2)n4n\sum_{n=0}^{\infty} \frac{(-1)^n (x-2)^n}{4^n}∑n=0∞4n(−1)n(x−2)n with interval (−2,6)(-2, 6)(−2,6)
∑n=0∞(x−2)n4n+1\sum_{n=0}^{\infty} \frac{(x-2)^n}{4^{n+1}}∑n=0∞4n+1(x−2)n with interval [−2,6)[-2, 6)[−2,6)
∑n=0∞(−1)n(x−2)n4n+1\sum_{n=0}^{\infty} \frac{(-1)^n (x-2)^n}{4^{n+1}}∑n=0∞4n+1(−1)n(x−2)n with interval [−2,6][-2, 6][−2,6]