Answer is: option3
\(2\)Solution:
To determine the radius of convergence, evaluate \[ \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right| \]
\[ \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right| = \lim_{k\to\infty} \left| \frac{(k+2)(x-3)^{k+1}}{(2k+3)2^{k+1}} \cdot \frac{(2k+1)2^k}{(k+1)(x-3)^k} \right| \]
\[ = \left|\frac{x-3}{2}\right| \lim_{k\to\infty} \frac{(k+2)(2k+1)}{(2k+3)(k+1)} \]
\[ = \left|\frac{x-3}{2}\right| \]
The series converges if \[ \left|\frac{x-3}{2}\right|<1 \]
Thus \[ |x-3|<2 \]
This gives an interval centered at \(x=3\) with radius \[ r=2 \]
