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(4x)=1/(1 - 4x) which of the following is the power series representation for f(4x)? a. ∑(k = 0 to ∞)x^(4k) b. ∑(k = 0 to ∞)4x^k c. ∑(k = 0 to ∞)1/(1 - (4x)^k) d. ∑(k = 0 to ∞)(4x)^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(4x)=1/(1 - 4x) which of the following is the power series representation for f(4x)? a. ∑(k = 0 to ∞)x^(4k) b. ∑(k = 0 to ∞)4x^k c. ∑(k = 0 to ∞)1/(1 - (4x)^k) d. ∑(k = 0 to ∞)(4x)^k the interval of convergence is . (simplify your answer. type your answer in interval notation.)

Answer

Explanation:

Step1: Substitute $u = 4x$ into geometric - series formula

We know that $\frac{1}{1 - u}=\sum_{k = 0}^{\infty}u^{k}$ for $|u|\lt1$. Let $u = 4x$, then $\frac{1}{1 - 4x}=\sum_{k = 0}^{\infty}(4x)^{k}$ for $|4x|\lt1$.

Step2: Find the interval of convergence

Solve the inequality $|4x|\lt1$. We can rewrite it as $- 1\lt4x\lt1$. Divide each part of the compound - inequality by 4: $-\frac{1}{4}\lt x\lt\frac{1}{4}$.

Answer:

D. $\sum_{k = 0}^{\infty}(4x)^{k}$ $(-\frac{1}{4},\frac{1}{4})$