# Find the Maclaurin series for f(x)=x^2/(1−8x^8). x^2/(1−8x^8)=sum_(n=0)^(oo) [_______________]

From Rogawski $ET$ $2e$ section $10.7$, exercise $4$.
Find the Maclaurin series for $f\left(x\right)=\frac{{x}^{2}}{1-8{x}^{8}}$.
$\frac{{x}^{2}}{1-8{x}^{8}}=\sum _{n=0}^{\mathrm{\infty }}\left[\text{_______________}\right]$
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tykoyz
Notice $\frac{1}{1-x}=\sum {x}^{n}$ . Hence
$\frac{{x}^{2}}{1-8{x}^{8}}={x}^{2}\sum \left(8{x}^{8}{\right)}^{n}=\sum {8}^{n}{x}^{8n+2}$
and this is valid for $|8{x}^{8}|<1$
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Hayley Bernard
What you have got is almost the right answer.
$\frac{{x}^{2}}{1-8{x}^{8}}={x}^{2}\sum _{n=0}^{\mathrm{\infty }}\left(8{x}^{8}{\right)}^{n}=\sum _{n=0}^{\mathrm{\infty }}{8}^{n}{x}^{8n+2}$
, which is valid if $|8{x}^{8}|<1$,hence, $|x|<\frac{1}{{8}^{1/8}}$