The coefficient of `x^(8)` in the expansion of `(1+x+x^(2)+x^(3))^(4)` is

Options

A. 30
B. 31
C. 32
D. 36

Answer

To find the coefficient of \( x^8 \) in the expansion of \( (1 + x + x^2 + x^3)^4 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (1 + x + x^2 + x^3)^4 \]
This can be rewritten as:
\[ (1 + x)^4 (1 + x^2)^4 \]

### Step 2: Expand using the Binomial Theorem
Using the Binomial Theorem, we can expand both \( (1 + x)^4 \) and \( (1 + x^2)^4 \).

For \( (1 + x)^4 \):
\[ (1 + x)^4 = \sum_{k=0}^{4} \binom{4}{k} x^k = \binom{4}{0} + \binom{4}{1} x + \binom{4}{2} x^2 + \binom{4}{3} x^3 + \binom{4}{4} x^4 \]
Calculating the coefficients:
\[ = 1 + 4x + 6x^2 + 4x^3 + x^4 \]

For \( (1 + x^2)^4 \):
\[ (1 + x^2)^4 = \sum_{j=0}^{4} \binom{4}{j} (x^2)^j = \binom{4}{0} + \binom{4}{1} x^2 + \binom{4}{2} x^4 + \binom{4}{3} x^6 + \binom{4}{4} x^8 \]
Calculating the coefficients:
\[ = 1 + 4x^2 + 6x^4 + 4x^6 + x^8 \]

### Step 3: Multiply the expansions
Now we need to multiply the two expansions:
\[ (1 + 4x + 6x^2 + 4x^3 + x^4)(1 + 4x^2 + 6x^4 + 4x^6 + x^8) \]

### Step 4: Identify terms that contribute to \( x^8 \)
We will identify the combinations of terms from both expansions that result in \( x^8 \):

1. **From \( 1 \) in \( (1 + x)^4 \) and \( x^8 \) in \( (1 + x^2)^4 \)**:
- Contribution: \( 1 \times 1 = 1 \)

2. **From \( 4x^2 \) in \( (1 + x)^4 \) and \( 6x^4 \) in \( (1 + x^2)^4 \)**:
- Contribution: \( 4 \times 6 = 24 \)

3. **From \( 6x^4 \) in \( (1 + x)^4 \) and \( 4x^2 \) in \( (1 + x^2)^4 \)**:
- Contribution: \( 6 \times 4 = 24 \)

4. **From \( 4x^3 \) in \( (1 + x)^4 \) and \( 4x^6 \) in \( (1 + x^2)^4 \)**:
- Contribution: \( 4 \times 0 = 0 \) (not applicable since \( x^6 \) does not contribute to \( x^8 \))

5. **From \( x^4 \) in \( (1 + x)^4 \) and \( 1 \) in \( (1 + x^2)^4 \)**:
- Contribution: \( 1 \times 0 = 0 \) (not applicable since \( x^4 \) does not contribute to \( x^8 \))

### Step 5: Sum the contributions
Now we sum the contributions:
\[ 1 + 24 + 24 = 49 \]

### Final Answer
Thus, the coefficient of \( x^8 \) in the expansion of \( (1 + x + x^2 + x^3)^4 \) is \( \boxed{49} \).