The coefficients of `x^(13)` in the expansion of
` (1 - x)^(5) (1 + x + x^(2) + x^(3) )^(4)` , is

Options

A. 24
B. 12
C. 6
D. 4

Answer

To find the coefficient of \( x^{13} \) in the expansion of \( (1 - x)^{5} (1 + x + x^{2} + x^{3})^{4} \), we can follow these steps:

### Step 1: Simplify the expression
We start with the expression:
\[ f(x) = (1 - x)^{5} \cdot (1 + x + x^{2} + x^{3})^{4} \]
The term \( (1 + x + x^{2} + x^{3}) \) can be rewritten using the formula for the sum of a geometric series:
\[ 1 + x + x^{2} + x^{3} = \frac{1 - x^{4}}{1 - x} \]
Thus, we can express \( f(x) \) as:
\[ f(x) = (1 - x)^{5} \cdot \left( \frac{1 - x^{4}}{1 - x} \right)^{4} \]

### Step 2: Expand the expression
Now, we can rewrite \( f(x) \):
\[ f(x) = (1 - x)^{5} \cdot \frac{(1 - x^{4})^{4}}{(1 - x)^{4}} = \frac{(1 - x)^{5} (1 - x^{4})^{4}}{(1 - x)^{4}} = (1 - x)^{1} (1 - x^{4})^{4} \]
This simplifies to:
\[ f(x) = (1 - x)(1 - x^{4})^{4} \]

### Step 3: Expand \( (1 - x^{4})^{4} \)
Using the binomial theorem, we expand \( (1 - x^{4})^{4} \):
\[ (1 - x^{4})^{4} = \sum_{k=0}^{4} \binom{4}{k} (-1)^{k} x^{4k} = 1 - 4x^{4} + 6x^{8} - 4x^{12} + x^{16} \]

### Step 4: Multiply by \( (1 - x) \)
Now we multiply this expansion by \( (1 - x) \):
\[ f(x) = (1 - x)(1 - 4x^{4} + 6x^{8} - 4x^{12} + x^{16}) \]
Distributing \( (1 - x) \):
\[ f(x) = 1 - 4x^{4} + 6x^{8} - 4x^{12} + x^{16} - x + 4x^{5} - 6x^{9} + 4x^{13} - x^{17} \]

### Step 5: Collect terms to find the coefficient of \( x^{13} \)
Now we identify the coefficient of \( x^{13} \):
- From \( -4x^{12} \) (coefficient is \( -4 \))
- From \( 4x^{5} \) (coefficient is \( 4 \))

Thus, the coefficient of \( x^{13} \) is:
\[ -4 + 4 = 0 \]

### Final Answer
The coefficient of \( x^{13} \) in the expansion of \( (1 - x)^{5} (1 + x + x^{2} + x^{3})^{4} \) is \( 0 \).