The coefficient of `x^4` in `(1+x+x^2+x^3)^11` is

Options

A. `-144`
B. 144
C. `990`
D. `-142`

Answer

To find the coefficient of \( x^4 \) in the expansion of \( (1 + x + x^2 + x^3)^{11} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (1 + x + x^2 + x^3)^{11} \]
This can be rewritten as:
\[ (1 + x^2)^{11} \cdot (1 + x)^{11} \]

### Step 2: Use the Binomial Theorem
According to the Binomial Theorem, we can expand \( (1 + x^2)^{11} \) and \( (1 + x)^{11} \):
\[ (1 + x^2)^{11} = \sum_{k=0}^{11} \binom{11}{k} (x^2)^k = \sum_{k=0}^{11} \binom{11}{k} x^{2k} \]
\[ (1 + x)^{11} = \sum_{m=0}^{11} \binom{11}{m} x^m \]

### Step 3: Identify combinations that yield \( x^4 \)
We need to find combinations of \( k \) and \( m \) such that:
\[ 2k + m = 4 \]
This gives us the following possible pairs of \( (k, m) \):
- \( k = 0, m = 4 \) (from \( x^0 \) in \( (1 + x^2)^{11} \) and \( x^4 \) in \( (1 + x)^{11} \))
- \( k = 1, m = 2 \) (from \( x^2 \) in \( (1 + x^2)^{11} \) and \( x^2 \) in \( (1 + x)^{11} \))
- \( k = 2, m = 0 \) (from \( x^4 \) in \( (1 + x^2)^{11} \) and \( x^0 \) in \( (1 + x)^{11} \))

### Step 4: Calculate the coefficients for each case
1. For \( k = 0, m = 4 \):
\[ \text{Coefficient} = \binom{11}{0} \cdot \binom{11}{4} = 1 \cdot 330 = 330 \]

2. For \( k = 1, m = 2 \):
\[ \text{Coefficient} = \binom{11}{1} \cdot \binom{11}{2} = 11 \cdot 55 = 605 \]

3. For \( k = 2, m = 0 \):
\[ \text{Coefficient} = \binom{11}{2} \cdot \binom{11}{0} = 55 \cdot 1 = 55 \]

### Step 5: Sum the coefficients
Now, we sum the coefficients from all cases:
\[ 330 + 605 + 55 = 990 \]

### Final Answer
Thus, the coefficient of \( x^4 \) in the expansion of \( (1 + x + x^2 + x^3)^{11} \) is:
\[ \boxed{990} \]