use the geometric series f(x)=1/(1 - x)=∑(k = 0 to ∞)x^k, for |x| < 1, to find the power series…

use the geometric series f(x)=1/(1 - x)=∑(k = 0 to ∞)x^k, for |x| < 1, to find the power series representation for the following function (centered at 0). give the interval of convergence of the new series. f(x^3)=1/(1 - x^3) which of the following is the power series representation for f(x^3)? a. ∑(k = 0 to ∞)3x^k b. ∑(k = 0 to ∞)1/(1 - x^3k) c. ∑(k = 0 to ∞)x^3k d. ∑(k = 0 to ∞)(3x)^k the interval of convergence is . (simplify your answer. type your answer in interval notation.)

use the geometric series f(x)=1/(1 - x)=∑(k = 0 to ∞)x^k, for |x| < 1, to find the power series representation for the following function (centered at 0). give the interval of convergence of the new series. f(x^3)=1/(1 - x^3) which of the following is the power series representation for f(x^3)? a. ∑(k = 0 to ∞)3x^k b. ∑(k = 0 to ∞)1/(1 - x^3k) c. ∑(k = 0 to ∞)x^3k d. ∑(k = 0 to ∞)(3x)^k the interval of convergence is . (simplify your answer. type your answer in interval notation.)

Answer

Explanation:

Step1: Substitute $x^3$ into geometric - series formula

Given the geometric series $\frac{1}{1 - x}=\sum_{k = 0}^{\infty}x^{k}$ for $|x|\lt1$. Substitute $u=x^{3}$ into the formula. Then $\frac{1}{1 - x^{3}}=\sum_{k = 0}^{\infty}(x^{3})^{k}=\sum_{k = 0}^{\infty}x^{3k}$.

Step2: Find the interval of convergence

Since the original geometric - series $\frac{1}{1 - u}=\sum_{k = 0}^{\infty}u^{k}$ converges for $|u|\lt1$, and $u = x^{3}$, we set $|x^{3}|\lt1$. Taking the cube - root of both sides of the inequality $|x^{3}|\lt1$, we get $|x|\lt1$. So the interval of convergence is $(-1,1)$.

Answer:

C. $\sum_{k = 0}^{\infty}x^{3k}$ $(-1,1)$