An archery target can be represented as three concentric circles with radii $3$, $2$, and $1$ which split the target into $3$ regions, as shown in the figure below. What is the area of Region $1$ plus the area of Region $3$?

[i]2015 CCA Math Bonanza Team Round #1[/i]
1. First, we need to find the area of each region. The regions are defined by the concentric circles with radii 3, 2, and 1.
2. The area of a circle is given by the formula \( A = \pi r^2 \).

3. Calculate the area of the largest circle (radius 3):
   \[
   A_{\text{large}} = \pi \times 3^2 = 9\pi
   \]

4. Calculate the area of the middle circle (radius 2):
   \[
   A_{\text{middle}} = \pi \times 2^2 = 4\pi
   \]

5. Calculate the area of the smallest circle (radius 1):
   \[
   A_{\text{small}} = \pi \times 1^2 = \pi
   \]

6. Region 1 is the area between the largest circle and the middle circle. Therefore, the area of Region 1 is:
   \[
   A_{\text{Region 1}} = A_{\text{large}} - A_{\text{middle}} = 9\pi - 4\pi = 5\pi
   \]

7. Region 3 is the area of the smallest circle. Therefore, the area of Region 3 is:
   \[
   A_{\text{Region 3}} = \pi
   \]

8. The total area of Region 1 plus the area of Region 3 is:
   \[
   A_{\text{Region 1}} + A_{\text{Region 3}} = 5\pi + \pi = 6\pi
   \]

The final answer is \(\boxed{6\pi}\)