To solve the problem, we need to find the coefficients \( a_0 \), \( a_1 \), and \( a_{30} \) from the expansion of \( (1 + 2x + 3x^2)^{15} \) and then determine the unit place of the sum \( a_0 + a_1 + a_{30} \).
### Step-by-Step Solution:
1. **Finding \( a_0 \)**:
- To find \( a_0 \), we substitute \( x = 0 \) in the expression \( (1 + 2x + 3x^2)^{15} \).
- This gives us:
\[
(1 + 2(0) + 3(0)^2)^{15} = 1^{15} = 1
\]
- Therefore, \( a_0 = 1 \).
**Hint**: The coefficient \( a_0 \) is the constant term in the expansion, which can be found by evaluating the expression at \( x = 0 \).
2. **Finding \( a_{30} \)**:
- To find \( a_{30} \), we need the coefficient of \( x^{30} \) in the expansion.
- The term \( (1 + 2x + 3x^2)^{15} \) can be thought of as choosing terms from \( 1 \), \( 2x \), and \( 3x^2 \) such that the total degree of \( x \) is 30.
- The only way to achieve \( x^{30} \) is to select \( 15 \) instances of \( 3x^2 \) (since \( 2x \) would only contribute to lower powers).
- Therefore, we calculate:
\[
a_{30} = \binom{15}{15} \cdot (3)^{15} = 1 \cdot 3^{15} = 3^{15}
\]
**Hint**: To find \( a_{30} \), consider how many times you can select \( 3x^2 \) to achieve \( x^{30} \) and use the binomial coefficient for the selection.
3. **Finding \( a_1 \)**:
- To find \( a_1 \), we differentiate the expression \( (1 + 2x + 3x^2)^{15} \) and then evaluate at \( x = 0 \).
- Differentiating gives:
\[
15(1 + 2x + 3x^2)^{14} \cdot (2 + 6x)
\]
- Evaluating at \( x = 0 \):
\[
15(1)^{14} \cdot (2) = 15 \cdot 2 = 30
\]
- Therefore, \( a_1 = 30 \).
**Hint**: The coefficient \( a_1 \) can be found by differentiating the entire expression and evaluating the result at \( x = 0 \).
4. **Calculating \( a_0 + a_1 + a_{30} \)**:
- Now we sum the coefficients:
\[
a_0 + a_1 + a_{30} = 1 + 30 + 3^{15}
\]
5. **Finding the unit place of \( 1 + 30 + 3^{15} \)**:
- First, calculate \( 1 + 30 = 31 \).
- Next, we need the unit place of \( 3^{15} \):
- The unit digits of powers of \( 3 \) cycle every 4: \( 3, 9, 7, 1 \).
- \( 3^1 \) has unit digit \( 3 \), \( 3^2 \) has unit digit \( 9 \), \( 3^3 \) has unit digit \( 7 \), \( 3^4 \) has unit digit \( 1 \).
- Since \( 15 \mod 4 = 3 \), the unit digit of \( 3^{15} \) is the same as that of \( 3^3 \), which is \( 7 \).
- Therefore, the unit place of \( 3^{15} \) is \( 7 \).
6. **Final Calculation**:
- Now we find the unit place of \( 31 + 3^{15} \):
\[
31 + 7 = 38
\]
- The unit place of \( 38 \) is \( 8 \).
### Conclusion:
The digit at the unit place of \( a_0 + a_1 + a_{30} \) is \( \boxed{8} \).
