Question

use substitution to find the taylor series at ( x = 0 ) of the function ( \frac{8}{1 - 9x} ) ( sum_{k = 0}^{infty}8cdot9^{k}x^{k} ) what is the taylor series for ( \frac{8}{1 - 9x} ) at ( x = 0 )? a. ( 72x+648x^{2}+5832x^{3}+52,488x^{4}+cdots ) b. ( 8 + 72x+648x^{2}+5832x^{3}+cdots ) c. ( 72x - 648x^{2}+5832x^{3}-52,488x^{4}+cdots ) d. ( 8 - 72x+648x^{2}-5832x^{3}+cdots )

Explanation:

Step1: Recall the geometric series formula

The geometric series formula is \(\sum_{k = 0}^{\infty}r^{k}=\frac{1}{1 - r}\), where \(|r|\lt1\).

Step2: Substitute \(r = 9x\) into the formula

We have \(\frac{1}{1-9x}=\sum_{k = 0}^{\infty}(9x)^{k}=\sum_{k = 0}^{\infty}9^{k}x^{k}\), for \(|9x|\lt1\) (i.e., \(|x|\lt\frac{1}{9}\)).

Step3: Multiply by 8

Multiply both sides of the equation \(\frac{1}{1 - 9x}=\sum_{k=0}^{\infty}9^{k}x^{k}\) by 8. So, \(\frac{8}{1 - 9x}=8\sum_{k = 0}^{\infty}9^{k}x^{k}=\sum_{k = 0}^{\infty}8\cdot9^{k}x^{k}\).
When \(k = 0\), the term is \(8\cdot9^{0}x^{0}=8\). When \(k = 1\), the term is \(8\cdot9^{1}x^{1}=72x\). When \(k = 2\), the term is \(8\cdot9^{2}x^{2}=648x^{2}\). When \(k = 3\), the term is \(8\cdot9^{3}x^{3}=5832x^{3}\).

Answer:

B. \(8 + 72x+648x^{2}+5832x^{3}+\cdots\)