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

Options

A. 4
B. -4
C. 9
D. none of these

Explanation

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:
\[ (1 - x)^{5} (1 + x + x^{2} + x^{3})^{4} \]
The term \( (1 + x + x^{2} + x^{3})^{4} \) 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 \( (1 + x + x^{2} + x^{3})^{4} \) as:
\[ \left( \frac{1 - x^{4}}{1 - x} \right)^{4} = (1 - x^{4})^{4} \cdot (1 - x)^{-4} \]

### Step 2: Rewrite the entire expression
Now we can rewrite the entire expression:
\[ (1 - x)^{5} \cdot (1 - x^{4})^{4} \cdot (1 - x)^{-4} \]
This simplifies to:
\[ (1 - x)^{1} \cdot (1 - x^{4})^{4} \]

### Step 3: Expand \( (1 - x^{4})^{4} \)
Using the Binomial Theorem, we expand \( (1 - x^{4})^{4} \):
\[ (1 - x^{4})^{4} = \sum_{k=0}^{4} \binom{4}{k} (-1)^{k} x^{4k} = \sum_{k=0}^{4} \binom{4}{k} (-1)^{k} x^{4k} \]

### Step 4: Combine with \( (1 - x)^{1} \)
Now we need to multiply this expansion by \( (1 - x)^{1} \):
\[ (1 - x) \cdot \left( \sum_{k=0}^{4} \binom{4}{k} (-1)^{k} x^{4k} \right) \]
This gives us:
\[ \sum_{k=0}^{4} \binom{4}{k} (-1)^{k} x^{4k} - \sum_{k=0}^{4} \binom{4}{k} (-1)^{k} x^{4k + 1} \]

### Step 5: Find the coefficient of \( x^{13} \)
To find the coefficient of \( x^{13} \), we need to consider:
1. From \( \sum_{k=0}^{4} \binom{4}{k} (-1)^{k} x^{4k} \), we need \( 4k = 13 \) which is not possible since \( k \) must be an integer.
2. From \( -\sum_{k=0}^{4} \binom{4}{k} (-1)^{k} x^{4k + 1} \), we need \( 4k + 1 = 13 \) which gives \( k = 3 \).

Thus, we need to find \( -\binom{4}{3} (-1)^{3} \):
\[ -\binom{4}{3} (-1)^{3} = -4 \cdot (-1) = 4 \]

### Final Answer
The coefficient of \( x^{13} \) in the expansion is \( 4 \).

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