Answer
$\Sigma_{n=0}^{\infty}(5)(4x^{2})^{n}$,
The interval of convergence is $(-\frac{1}{2},\frac{1}{2})$
The radius of convergence is $\frac{1}{2}$
Work Step by Step
The sum of a geometric series with initial term $a$ and common ratio $r$ is
$S=\Sigma_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$
Therefore,
$f(x)=\Sigma_{n=0}^{\infty}ar^{n}=\Sigma_{n=0}^{\infty}(5)(4x^{2})^{n}$
This is the power series representation of $f(x)$.
We know that the power series converges when $r=|4x^{2}|\lt 1$
$x^{2}\lt \frac{1}{4}$
Hence, the interval of convergence is $(-\frac{1}{2},\frac{1}{2})$
The radius of convergence is $\frac{1}{2}$
