Let `aa n db` be the coefficients of `x^3` in `(1+x+2x^2+3x^2)^4a n d(1+x+2x^2+3x^3+4x^4)^4,` then respectively. Then the value of `4a//b` is.

Answer

To solve the problem, we need to find the coefficients \( a \) and \( b \) of \( x^3 \) in the expressions \( (1+x+2x^2+3x^3)^4 \) and \( (1+x+2x^2+3x^3+4x^4)^4 \) respectively. Then, we will calculate the value of \( \frac{4a}{b} \).

### Step 1: Find the coefficient \( a \) in \( (1+x+2x^2+3x^3)^4 \)

We can use the multinomial expansion to find the coefficient of \( x^3 \).

The general term in the expansion of \( (1+x+2x^2+3x^3)^4 \) can be expressed as:

\[ \frac{4!}{k_1! k_2! k_3! k_4!} \cdot 1^{k_1} \cdot x^{k_2} \cdot (2x^2)^{k_3} \cdot (3x^3)^{k_4} \]

where \( k_1 + k_2 + k_3 + k_4 = 4 \) and \( k_2 + 2k_3 + 3k_4 = 3 \).

### Step 2: Find combinations of \( k_2, k_3, k_4 \)

We will consider the possible values of \( k_4 \) (the number of \( 3x^3 \) terms):

- **Case 1:** \( k_4 = 1 \)
- Then \( k_2 + 2k_3 = 0 \) implies \( k_2 = 0, k_3 = 0 \).
- Contribution: \( \frac{4!}{3!1!} \cdot 3^1 = 4 \cdot 3 = 12 \).

- **Case 2:** \( k_4 = 0 \)
- Then \( k_2 + 2k_3 = 3 \).
- Possible combinations:
- \( k_2 = 3, k_3 = 0 \): Contribution: \( \frac{4!}{1!3!0!} \cdot 2^0 = 4 \).
- \( k_2 = 1, k_3 = 1 \): Contribution: \( \frac{4!}{2!1!1!} \cdot 2^1 = 12 \).
- \( k_2 = 0, k_3 = 1 \): Contribution: \( \frac{4!}{1!0!2!} \cdot 2^2 = 12 \).

Adding all contributions together for \( a \):

\[ a = 12 + 4 + 12 + 12 = 40 \]

### Step 3: Find the coefficient \( b \) in \( (1+x+2x^2+3x^3+4x^4)^4 \)

We will again use the multinomial expansion for this expression.

The general term can be expressed similarly:

\[ \frac{4!}{k_1! k_2! k_3! k_4! k_5!} \cdot 1^{k_1} \cdot x^{k_2} \cdot (2x^2)^{k_3} \cdot (3x^3)^{k_4} \cdot (4x^4)^{k_5} \]

where \( k_1 + k_2 + k_3 + k_4 + k_5 = 4 \) and \( k_2 + 2k_3 + 3k_4 = 3 \).

### Step 4: Find combinations of \( k_2, k_3, k_4, k_5 \)

- **Case 1:** \( k_4 = 1 \)
- Then \( k_2 + 2k_3 + 0 = 0 \) implies \( k_2 = 0, k_3 = 0, k_5 = 3 \).
- Contribution: \( \frac{4!}{3!1!} \cdot 3^1 = 12 \).

- **Case 2:** \( k_4 = 0 \)
- Then \( k_2 + 2k_3 = 3 \).
- Possible combinations:
- \( k_2 = 3, k_3 = 0 \): Contribution: \( \frac{4!}{1!3!0!} \cdot 2^0 = 4 \).
- \( k_2 = 1, k_3 = 1 \): Contribution: \( \frac{4!}{2!1!1!} \cdot 2^1 = 12 \).
- \( k_2 = 0, k_3 = 1 \): Contribution: \( \frac{4!}{1!0!2!1!} \cdot 2^2 = 12 \).
- \( k_5 = 1 \): Contribution: \( \frac{4!}{1!0!2!1!} \cdot 4^1 = 24 \).

Adding all contributions together for \( b \):

\[ b = 12 + 4 + 12 + 12 + 24 = 64 \]

### Step 5: Calculate \( \frac{4a}{b} \)

Now, we can find:

\[ \frac{4a}{b} = \frac{4 \times 40}{64} = \frac{160}{64} = 2.5 \]

### Final Answer

Thus, the value of \( \frac{4a}{b} \) is \( 2.5 \).