Answer:
The answer is "[tex]a_n =(\frac{x^{8n+5}}{a^{8n+8}})[/tex]"
Step-by-step explanation:
[tex]f(x) =\frac{x^5}{a^8 -x^8} = \frac{\frac{x^5}{a^8}}{1-(\frac{x}{a})^8}\\\\\\\to \frac{1}{1-(\frac{x}{a})^8}[/tex]
The above equation can be writtern as the sum of the [tex]\infty[/tex] geometric series with [tex]r= \frac{x}{a}[/tex]
[tex]= \frac{\frac{x^5}{a^8}}{1+(\frac{x}{a})^8} \\\\ = \frac{x^5}{a^8} [ 1+ (\frac{x}{a})^8 + ((\frac{x}{a})^8)^2+.......]\\\\= \frac{x^5}{a^8} \sum_{\infty}^{n=0} (\frac{x}{a})^{8n}\\\\ = \sum_{\infty}^{n=0} (\frac{x^{8n+5}}{a^{8n+8}})\\\\a_n =(\frac{x^{8n+5}}{a^{8n+8}})[/tex]
