Answer is: option1
\(3\)Solution:
Let \(R\) denote the ratio of successive terms: \[ \left|\frac{a_{n+1}}{a_n}\right| \]
Then \[ \left|\frac{a_{n+1}}{a_n}\right| = \left| \frac{(-1)^{n+2}(x-2)^{n+1}}{(n+1)3^{n+1}} \cdot \frac{n3^n}{(-1)^{n+1}(x-2)^n} \right| \]
\[ = \left| \frac{(-1)(x-2)}{3}\cdot\frac{n}{n+1} \right| \]
Hence \[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x-2}{3}\right| \]
For convergence, we must have \[ \left|\frac{x-2}{3}\right|<1 \] so that \[ |x-2|<3 \]
Hence the radius of convergence is \[ r=3 \]
