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

Options

A. `""^(n)C_(4)`
B. `""^(n)C_(4)+""^(n)C_(2)`
C. `""^(n)C_(4)+""^(n)C_(1)+""^(n)C_(4).""^(n)C_(2)`
D. `""^(n)C_(4)+""^(n)C_(2)+""^(n)C_(1).""^(n)C_(2)`

Answer

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

### Step 1: Factor the expression
We can rewrite the expression \( 1 + x + x^2 + x^3 \) as \( (1 + x)(1 + x^2) \). Thus, we have:

\[ (1 + x + x^2 + x^3)^n = (1 + x)(1 + x^2)^n \]

### Step 2: Expand the expression
Now we will expand \( (1 + x)^n \) and \( (1 + x^2)^n \) using the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

Applying this, we have:

\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

\[ (1 + x^2)^n = \sum_{j=0}^{n} \binom{n}{j} (x^2)^j = \sum_{j=0}^{n} \binom{n}{j} x^{2j} \]

### Step 3: Identify the combinations that yield \( x^4 \)
To find the coefficient of \( x^4 \) in the product of these two expansions, we need to consider the pairs \( (k, j) \) such that:

\[ k + 2j = 4 \]

This gives us the following cases:

1. \( k = 4, j = 0 \)
2. \( k = 2, j = 1 \)
3. \( k = 0, j = 2 \)

### Step 4: Calculate the coefficients for each case
Now we calculate the coefficients for each case:

1. For \( k = 4, j = 0 \):
\[ \text{Coefficient} = \binom{n}{4} \cdot \binom{n}{0} = \binom{n}{4} \]

2. For \( k = 2, j = 1 \):
\[ \text{Coefficient} = \binom{n}{2} \cdot \binom{n}{1} = \binom{n}{2} \cdot n \]

3. For \( k = 0, j = 2 \):
\[ \text{Coefficient} = \binom{n}{0} \cdot \binom{n}{2} = \binom{n}{2} \]

### Step 5: Combine the coefficients
Now, we combine all the coefficients to find the total coefficient of \( x^4 \):

\[ \text{Total Coefficient} = \binom{n}{4} + n \cdot \binom{n}{2} + \binom{n}{2} \]

This simplifies to:

\[ \text{Total Coefficient} = \binom{n}{4} + (n + 1) \cdot \binom{n}{2} \]

### Final Answer
Thus, the coefficient of \( x^4 \) in the expansion of \( (1 + x + x^2 + x^3)^n \) is:

\[ \binom{n}{4} + (n + 1) \cdot \binom{n}{2} \]