Find the coefficient of `x^4` in the expansion of `(1+x+x^2+x^3)^6`

Options

A. 100
B. 110
C. 120
D. 125

Answer

To find the coefficient of \( x^4 \) in the expansion of \( (1 + x + x^2 + x^3)^6 \), we can follow these steps:

### Step 1: Rewrite the expression
The expression can be rewritten as:
\[ (1 + x + x^2 + x^3)^6 = (1 + x(1 + x + x^2))^6 \]
However, a more straightforward approach is to recognize that \( 1 + x + x^2 + x^3 \) can be expressed in a more manageable form.

### Step 2: Factor the expression
Notice that:
\[ 1 + x + x^2 + x^3 = \frac{1 - x^4}{1 - x} \quad \text{for } |x| < 1 \]
Thus, we can write:
\[ (1 + x + x^2 + x^3)^6 = \left(\frac{1 - x^4}{1 - x}\right)^6 = (1 - x^4)^6 (1 - x)^{-6} \]

### Step 3: Expand using the Binomial Theorem
Now we can expand both parts using the Binomial Theorem.

1. **Expanding \( (1 - x^4)^6 \)**:
\[ (1 - x^4)^6 = \sum_{k=0}^{6} \binom{6}{k} (-1)^k x^{4k} \]

2. **Expanding \( (1 - x)^{-6} \)**:
\[ (1 - x)^{-6} = \sum_{m=0}^{\infty} \binom{m + 5}{5} x^m \]

### Step 4: Find the coefficient of \( x^4 \)
To find the coefficient of \( x^4 \) in the product of these two expansions, we need to consider the contributions from different values of \( k \) from the first expansion:

- For \( k = 0 \): We get \( \binom{6}{0} (-1)^0 x^0 \) from \( (1 - x^4)^6 \) and need \( x^4 \) from \( (1 - x)^{-6} \):
- Coefficient: \( \binom{4 + 5}{5} = \binom{9}{5} = 126 \)

- For \( k = 1 \): We get \( \binom{6}{1} (-1)^1 x^4 \) from \( (1 - x^4)^6 \) and need \( x^0 \) from \( (1 - x)^{-6} \):
- Coefficient: \( -\binom{6}{1} = -6 \)

- For \( k = 2 \): We get \( \binom{6}{2} (-1)^2 x^8 \) from \( (1 - x^4)^6 \) but we need \( x^{-4} \) from \( (1 - x)^{-6} \), which does not contribute.

### Step 5: Combine the contributions
Now, we combine the contributions:
\[ \text{Coefficient of } x^4 = 126 - 6 = 120 \]

### Final Answer
Thus, the coefficient of \( x^4 \) in the expansion of \( (1 + x + x^2 + x^3)^6 \) is \( \boxed{120} \).