Given series is

\(\sum_{n=0}^\infty\frac{(x-4)^{2n}}{36^n}\)

Now,

let \((x-4)^2=y\)

So, the series becomes

\(\sum_{n=0}^\infty\frac{y^n}{36^n}\)

So, \(a_n=\frac{1}{36^n}\)

Then,

\((a_n)^\frac{1}{n}=\frac{1}{36}\)

and

\(\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=\frac{1}{36}\)

So, radius of convergence is 36

Then, the radius of convergence of the given series is 6

Interval of convergence \(=(-6-4,-6+4)=(-10,-2)\)

Now,

\(a_0=1\)

\(a_1=\frac{y}{36}\)

\(a_2=\frac{y^2}{36^2}\)

\(\vdots\)

\(a_n=\frac{y^n}{36^n}\)

So, \(S_n=a_0+a_1+...+a_n\)

\(=1+\frac{y}{36}+\frac{y^2}{36^2}+...+\frac{y^n}{36^n}\)

\(=\frac{1[1-(\frac{y}{36})^n]}{1-\frac{y}{36}}\)

\(=\frac{36[1-\frac{(x-4)^{2n}}{36^n}]}{36-(x-4)^2}\)

\(=\frac{36^n-(x-4)^{2n}}{36^{n-1}[36-(x-4)^2]}\)

Hence, the sum of the series as a function of x is

\(=\frac{36^n-(x-4)^{2n}}{36^{n-1}[36-(x-4)^2]}\)

\(\sum_{n=0}^\infty\frac{(x-4)^{2n}}{36^n}\)

Now,

let \((x-4)^2=y\)

So, the series becomes

\(\sum_{n=0}^\infty\frac{y^n}{36^n}\)

So, \(a_n=\frac{1}{36^n}\)

Then,

\((a_n)^\frac{1}{n}=\frac{1}{36}\)

and

\(\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=\frac{1}{36}\)

So, radius of convergence is 36

Then, the radius of convergence of the given series is 6

Interval of convergence \(=(-6-4,-6+4)=(-10,-2)\)

Now,

\(a_0=1\)

\(a_1=\frac{y}{36}\)

\(a_2=\frac{y^2}{36^2}\)

\(\vdots\)

\(a_n=\frac{y^n}{36^n}\)

So, \(S_n=a_0+a_1+...+a_n\)

\(=1+\frac{y}{36}+\frac{y^2}{36^2}+...+\frac{y^n}{36^n}\)

\(=\frac{1[1-(\frac{y}{36})^n]}{1-\frac{y}{36}}\)

\(=\frac{36[1-\frac{(x-4)^{2n}}{36^n}]}{36-(x-4)^2}\)

\(=\frac{36^n-(x-4)^{2n}}{36^{n-1}[36-(x-4)^2]}\)

Hence, the sum of the series as a function of x is

\(=\frac{36^n-(x-4)^{2n}}{36^{n-1}[36-(x-4)^2]}\)