If `(1+2x+3x^(2))^(10)=a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+ . . .+a_(20)x^(20),` then

Options

A. `a_(1)=20`
B. `a_(2)=210`
C. `a_(4)=8085`
D. `a_(20)=2^(2).3^(7).7`

Answer

To solve the problem of finding the coefficients \( a_1, a_2, a_4, \) and \( a_{20} \) in the expansion of \( (1 + 2x + 3x^2)^{10} \), we can use the multinomial theorem.

### Step-by-Step Solution:

1. **Understanding the Expansion**:
We need to expand \( (1 + 2x + 3x^2)^{10} \) and find the coefficients of \( x^1, x^2, x^4, \) and \( x^{20} \).

2. **Finding \( a_{20} \)**:
The term \( x^{20} \) can only be formed by taking \( 3x^2 \) ten times, since \( 2x \) and \( 1 \) cannot contribute to \( x^{20} \). Therefore, we have:
\[ a_{20} = \text{Coefficient of } (3x^2)^{10} = 3^{10} \]
\[ a_{20} = 59049 \]

3. **Finding \( a_1 \)**:
To find \( a_1 \), we need to select \( x \) from \( 2x \) once and \( 1 \) from the remaining 9 terms:
\[ a_1 = \binom{10}{1} \cdot (2) \cdot (1)^{9} = 10 \cdot 2 = 20 \]

4. **Finding \( a_2 \)**:
For \( a_2 \), we can have:
- \( x^2 \) from \( 3x^2 \) once and \( 1 \) from the remaining 9 terms.
- \( x \) from \( 2x \) twice and \( 1 \) from the remaining 8 terms.

Therefore,
\[ a_2 = \binom{10}{1} \cdot 3 \cdot 1^8 + \binom{10}{2} \cdot (2^2) \cdot 1^8 \]
\[ = 10 \cdot 3 + 45 \cdot 4 = 30 + 180 = 210 \]

5. **Finding \( a_4 \)**:
For \( a_4 \), we can have:
- \( x^2 \) from \( 3x^2 \) twice and \( 1 \) from the remaining 8 terms.
- \( x \) from \( 2x \) four times and \( 1 \) from the remaining 6 terms.

Therefore,
\[ a_4 = \binom{10}{2} \cdot 3^2 \cdot 1^8 + \binom{10}{4} \cdot (2^4) \cdot 1^6 \]
\[ = 45 \cdot 9 + 210 \cdot 16 = 405 + 3360 = 3965 \]

### Final Coefficients:
- \( a_1 = 20 \)
- \( a_2 = 210 \)
- \( a_4 = 3965 \)
- \( a_{20} = 59049 \)

### Summary:
The coefficients are:
- \( a_1 = 20 \)
- \( a_2 = 210 \)
- \( a_4 = 3965 \)
- \( a_{20} = 59049 \)