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

Options

A. `a_(1) =20`
B. `a_(2) =210`
C. `a_(4) =8085`
D. `a_(20) = 2^(2) xx3^(7) xx7`.

Answer

To solve the problem, we need to find the coefficients \( a_1 \), \( a_2 \), and \( a_4 \) from the expansion of \( (1 + 2x + 3x^2)^{10} \).

### Step 1: Identify the expression
We start with the expression:
\[ (1 + 2x + 3x^2)^{10} \]

### Step 2: Use the multinomial theorem
The multinomial theorem states that:
\[ (a_1 + a_2 + a_3)^n = \sum_{k_1 + k_2 + k_3 = n} \frac{n!}{k_1! k_2! k_3!} a_1^{k_1} a_2^{k_2} a_3^{k_3} \]
In our case, \( a_1 = 1 \), \( a_2 = 2x \), and \( a_3 = 3x^2 \).

### Step 3: Find \( a_1 \)
To find \( a_1 \), we need the coefficient of \( x^1 \). This occurs when \( k_2 = 1 \) (from \( 2x \)) and \( k_1 + k_3 = 9 \). The number of ways to choose \( k_1 \) and \( k_3 \) such that \( k_1 + k_3 = 9 \) can be calculated as follows:
\[ \text{Coefficient of } x^1 = \frac{10!}{9!1!} (2)^1 (3)^0 = 10 \cdot 2 = 20 \]
Thus, \( a_1 = 20 \).

### Step 4: Find \( a_2 \)
To find \( a_2 \), we need the coefficient of \( x^2 \). This can occur in two ways:
1. \( k_2 = 0 \) and \( k_3 = 1 \) (from \( 3x^2 \)).
2. \( k_2 = 1 \) and \( k_1 = 8 \) (from \( 2x \) and \( 1 \)).

Calculating both cases:
1. For \( k_3 = 1 \) and \( k_1 = 9 \):
\[ \text{Coefficient} = \frac{10!}{9!0!1!} (1)^9 (3)^1 = 10 \cdot 3 = 30 \]

2. For \( k_2 = 1 \) and \( k_1 = 8 \):
\[ \text{Coefficient} = \frac{10!}{8!1!1!} (2)^1 (1)^8 = 90 \cdot 2 = 180 \]

Adding both contributions:
\[ a_2 = 30 + 180 = 210 \]

### Step 5: Find \( a_4 \)
To find \( a_4 \), we need the coefficient of \( x^4 \). This can occur in several ways:
1. \( k_3 = 2 \) and \( k_2 = 0 \) (from \( 3x^2 \)).
2. \( k_3 = 1 \) and \( k_2 = 2 \).
3. \( k_3 = 0 \) and \( k_2 = 4 \).

Calculating:
1. For \( k_3 = 2 \) and \( k_1 = 8 \):
\[ \text{Coefficient} = \frac{10!}{8!0!2!} (1)^8 (3)^2 = 45 \cdot 9 = 405 \]

2. For \( k_3 = 1 \) and \( k_2 = 2 \):
\[ \text{Coefficient} = \frac{10!}{7!2!1!} (1)^7 (2)^2 (3)^1 = 120 \cdot 4 \cdot 3 = 1440 \]

3. For \( k_3 = 0 \) and \( k_2 = 4 \):
\[ \text{Coefficient} = \frac{10!}{6!4!0!} (1)^6 (2)^4 = 210 \cdot 16 = 3360 \]

Adding all contributions:
\[ a_4 = 405 + 1440 + 3360 = 5205 \]

### Conclusion
The coefficients are:
- \( a_1 = 20 \)
- \( a_2 = 210 \)
- \( a_4 = 5205 \)