We are tasked with determining the third-degree Taylor polynomial for the function $f$ about $x = 2$. The function has derivatives of all orders, and we are provided with the formula for the $n$-th derivative of $f$ at $x = 2$:

$f^{(n)}(2) = \frac{n!}{n \cdot 3^n}$

Steps:

1. Calculate the derivatives at $x = 2$:

   • $f(2) = 0$ (given)
   • $f'(2) = \frac{1!}{1 \cdot 3^1} = \frac{1}{3}$
   • $f''(2) = \frac{2!}{2 \cdot 3^2} = \frac{2 \cdot 1}{2 \cdot 9} = \frac{1}{9}$
   • $f^{(3)}(2) = \frac{3!}{3 \cdot 3^3} = \frac{6}{3 \cdot 27} = \frac{2}{27}$

2. Construct the third-degree Taylor polynomial: The formula for the third-degree Taylor polynomial $T_3(x)$ centered at $x = 2$ is:

   $T_3(x) = f(2) + f'(2)(x - 2) + \frac{f''(2)}{2!}(x - 2)^2 + \frac{f^{(3)}(2)}{3!}(x - 2)^3$

   Substituting the calculated values:

   $T_3(x) = 0 + \frac{1}{3}(x - 2) + \frac{1}{9 \cdot 2}(x - 2)^2 + \frac{2}{27 \cdot 6}(x - 2)^3$

   Simplifying the coefficients:

   $T_3(x) = \frac{1}{3}(x - 2) + \frac{1}{18}(x - 2)^2 + \frac{1}{81}(x - 2)^3$

3. Match with the given options: This matches option B.

Thus, the correct answer is B.

Would you like more details or have any questions? Here are five additional related questions:

1. What is the general formula for a Taylor series expansion?
2. How do you find the radius of convergence for a Taylor series?
3. Why is it important that a function has derivatives of all orders for a Taylor series?
4. What does the remainder term in the Taylor series tell us about the accuracy of the approximation?
5. How does shifting the center of expansion (e.g., from $x = 0$ to $x = 2$) affect the Taylor polynomial?

Tip: Remember that higher-order terms in a Taylor expansion provide better approximations near the center of expansion.