Let $f_n(x)=\underbrace{xx\cdots x}_{n\ \text{times}}$ that is, $f_n(x)$ is a $n$ digit number with all digits $x$, where $x\in \{1,2,\cdots,9\}$. Then which of the following is $\Big(f_n(3)\Big)^2+f_n(2)$?
[list=1]
[*] $f_n(5)$
[*] $f_{2n}(9)$
[*] $f_{2n}(1)$
[*] None of these
[/list]
1. First, let's understand the function \( f_n(x) \). The function \( f_n(x) \) represents an \( n \)-digit number where each digit is \( x \). For example, \( f_3(3) = 333 \).

2. We can express \( f_n(x) \) in terms of powers of 10. Specifically, \( f_n(x) = x \cdot \frac{10^n - 1}{9} \). This is because the number \( 111 \ldots 1 \) (with \( n \) ones) can be written as \( \frac{10^n - 1}{9} \).

3. Therefore, \( f_n(3) = 3 \cdot \frac{10^n - 1}{9} \) and \( f_n(2) = 2 \cdot \frac{10^n - 1}{9} \).

4. We need to find \( \left( f_n(3) \right)^2 + f_n(2) \):
   \[
   \left( f_n(3) \right)^2 + f_n(2) = \left( 3 \cdot \frac{10^n - 1}{9} \right)^2 + 2 \cdot \frac{10^n - 1}{9}
   \]

5. Simplify \( \left( 3 \cdot \frac{10^n - 1}{9} \right)^2 \):
   \[
   \left( 3 \cdot \frac{10^n - 1}{9} \right)^2 = 9 \cdot \left( \frac{10^n - 1}{9} \right)^2 = 9 \cdot \frac{(10^n - 1)^2}{81} = \frac{(10^n - 1)^2}{9}
   \]

6. Now, add \( f_n(2) \):
   \[
   \left( f_n(3) \right)^2 + f_n(2) = \frac{(10^n - 1)^2}{9} + 2 \cdot \frac{10^n - 1}{9}
   \]

7. Combine the terms over a common denominator:
   \[
   \left( f_n(3) \right)^2 + f_n(2) = \frac{(10^n - 1)^2 + 2(10^n - 1)}{9}
   \]

8. Simplify the numerator:
   \[
   (10^n - 1)^2 + 2(10^n - 1) = 10^{2n} - 2 \cdot 10^n + 1 + 2 \cdot 10^n - 2 = 10^{2n} - 1
   \]

9. Therefore:
   \[
   \left( f_n(3) \right)^2 + f_n(2) = \frac{10^{2n} - 1}{9}
   \]

10. Notice that \( \frac{10^{2n} - 1}{9} \) is exactly \( f_{2n}(1) \), which is a \( 2n \)-digit number with all digits being 1.

Conclusion:
\[
\left( f_n(3) \right)^2 + f_n(2) = f_{2n}(1)
\]

The final answer is \(\boxed{f_{2n}(1)}\)