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The problem is asking for the second derivative of the function \( f(x) \) at \( x = 2 \), using the given Taylor series expansion:

\[
f(x) = \sum_{n=0}^{\infty} 3^{2n} (x - 2)^n
\]

This is a power series expansion of \( f(x) \) centered at \( a = 2 \). The general form of the Taylor series for \( f(x) \) around \( a = 2 \) is:

\[
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!} (x - 2)^n
\]

From the series given, we compare terms:

\[
f(x) = 3^0 + 3^2 (x-2)^1 + 3^4 (x-2)^2 + \dots
\]

The coefficient of \( (x-2)^n \) gives us \( \frac{f^{(n)}(2)}{n!} \), so:

- The constant term (\( (x-2)^0 \)) is \( 1 \), meaning \( f(2) = 1 \).
- The coefficient of \( (x-2)^1 \) is \( 3^2 = 9 \), so \( f'(2) = 9 \).
- The coefficient of \( (x-2)^2 \) is \( 3^4 = 81 \), and since \( \frac{f^{(2)}(2)}{2!} = 81 \), we solve for \( f^{(2)}(2) \):

\[
\frac{f^{(2)}(2)}{2!} = 81 \quad \Rightarrow \quad \frac{f^{(2)}(2)}{2} = 81 \quad \Rightarrow \quad f^{(2)}(2) = 81 \times 2 = 162
\]

So, the correct answer is \( f^{(2)}(2) = 162 \), which corresponds to option D, not B. 

Would you like more details on Taylor series or further explanation?

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Here are 5 related questions to explore:

1. What is the general formula for the \(n\)-th derivative in a Taylor series expansion?
2. How do you determine the radius of convergence for a power series?
3. Can you apply the Taylor series for functions centered at points other than zero?
4. What is the role of factorial terms in a Taylor series?
5. How do we interpret the coefficients in a Taylor series?

**Tip**: The coefficients in a Taylor series tell us a lot about the behavior of derivatives at the expansion point!

The Taylor series expansion of f(x) is \( \sum_{n=0}^{\infty} 3^{2n}(x-2)^n \) at a = 2. Find \( f^{(2)}(2) \).

Learn how to find the second derivative of a function using its Taylor series expansion. This problem involves calculating the second derivative at x=2 for the function f(x) given by a power series expansion.

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