find the series interval of convergence and, within this interval, the sum of the series as a function of…

find the series interval of convergence and, within this interval, the sum of the series as a function of x\n\n sum _ { n = 0 } ^ { infty } 16 ^ { n } x ^ { n } \n\nthe interval of convergence is ( left( - \frac { 1 } { 10 } \frac { 1 } { 10 } \right) )\n(simplify your answer type your answer in interval notation )\n\nwithin the interval ( sum _ { n = 0 } ^ { infty } 16 ^ { n } x ^ { n } = )

find the series interval of convergence and, within this interval, the sum of the series as a function of x\n\n sum _ { n = 0 } ^ { infty } 16 ^ { n } x ^ { n } \n\nthe interval of convergence is ( left( - \frac { 1 } { 10 } \frac { 1 } { 10 } \right) )\n(simplify your answer type your answer in interval notation )\n\nwithin the interval ( sum _ { n = 0 } ^ { infty } 16 ^ { n } x ^ { n } = )

Answer

Explanation:

Step1: Recognize the geometric series form

The series (\sum_{n = 0}^{\infty}16^{n}x^{n}=\sum_{n = 0}^{\infty}(16x)^{n}). A geometric series has the form (\sum_{n=0}^{\infty}r^{n}), where (r = 16x) in this case.

Step2: Use the formula for the sum of a geometric series

The sum of a geometric series (\sum_{n = 0}^{\infty}r^{n}=\frac{1}{1 - r}), when (|r|\lt1). Substituting (r = 16x) into the formula, we get (\frac{1}{1-16x}), since for convergence (|16x|\lt1) (which gives the interval (-\frac{1}{10}\lt x\lt\frac{1}{10}) as already found for convergence).

Answer:

(\frac{1}{1 - 16x})