Given series is

\(\sum_{n=1}^\infty\frac{(2x+9)^n}{n^2}\)

\(a_n=\frac{(2x+9)^n}{n^2}\)

We will use Ratio Test to find radius of convergence of the series.

\(\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\to\infty}|\frac{(2x+9)^{n+1}}{(n+1)^2}\cdot\frac{n^2}{(2x+9)^n}|=\lim_{n\to\infty}|(2x+9)\frac{n^2}{(2x+9)^n}|=|2x+9|\)

By Ratio Test , given series converges for

\(|2x+9|<1\)

\(2|x-(\frac{9}{2})|<1\)

\(|x-(\frac{9}{2})|<\frac{1}{2}\)

Therefore, radius of convergence is \(R=\frac{1}{2}\)

Now,

\(|2x+9|<1\)

\(-1<2x+9<1\)

\(-1-9<2x<1-9\)

\(-10<2x<-8\)

\(-5<x<-4\)

When \(x=5\), series become

\(\sum_{n=1}^\infty\frac{(2(-5)+9)^n}{n^2}=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\)

Which is convergent by Alternating series test ( because sequence \(\frac{1}{n^2}\) is decreasing and converging to 0 )

When \(x=-4\), series become

\(\sum_{n=1}^\infty\frac{(2(-4)+9)^n}{n^2}=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=\sum_{n=1}^\infty\frac{1}{n^2}\)

Which is also convergent by p-series test.

Hence, interval of convergence is \([-5,-4]\)

And:

Radius of convergence is \(R=\frac{1}{2}\)

Interval of convergence is \([-5,-4]\)