Question

use substitution to find the taylor series at ( x = 0 ) of the function ( \frac{8}{1 - 9x} ).
what is the general expression for the kth term for the taylor series at ( x = 0 ) for ( \frac{8}{1 - 9x} )?
( sum_{k = 0}^{infty}square )

Explanation:

Step1: Recall the geometric series formula

The geometric series formula is \(\frac{1}{1 - t}=\sum_{k = 0}^{\infty}t^{k}\), for \(|t|<1\).

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

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

Step3: Multiply by 8

Multiply both sides of the equation \(\frac{1}{1 - 9x}=\sum_{k=0}^{\infty}(9x)^{k}\) by 8. So, \(\frac{8}{1-9x}=8\sum_{k = 0}^{\infty}(9x)^{k}=\sum_{k = 0}^{\infty}8\times9^{k}x^{k}\).

Answer:

\(\sum_{k = 0}^{\infty}8\times9^{k}x^{k}\)