The coefficient of `x^(256)` in the expansion of `(1-x)^(101) (x^(2) + x + 1)^(100)` is :

Options

A. `""^(100)C_(16)`
B. `""^(100)C_(15)`
C. `-""^(100)C_(16)`
D. `-""^(100)C_(15)`

Answer

To find the coefficient of \( x^{256} \) in the expansion of \( (1 - x)^{101} (x^2 + x + 1)^{100} \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ (1 - x)^{101} (x^2 + x + 1)^{100} \]
We can rewrite \( (x^2 + x + 1)^{100} \) using the identity for the sum of cubes:
\[ x^2 + x + 1 = \frac{1 - x^3}{1 - x} \]
Thus, we can express \( (x^2 + x + 1)^{100} \) as:
\[ (x^2 + x + 1)^{100} = \left(\frac{1 - x^3}{1 - x}\right)^{100} = (1 - x^3)^{100} (1 - x)^{-100} \]

### Step 2: Substitute Back into the Expression
Substituting this back into our original expression gives:
\[ (1 - x)^{101} (1 - x^3)^{100} (1 - x)^{-100} \]
This simplifies to:
\[ (1 - x)^{1} (1 - x^3)^{100} = (1 - x)(1 - x^3)^{100} \]

### Step 3: Expand \( (1 - x^3)^{100} \)
Using the binomial theorem, we can expand \( (1 - x^3)^{100} \):
\[ (1 - x^3)^{100} = \sum_{k=0}^{100} \binom{100}{k} (-1)^k x^{3k} \]

### Step 4: Combine with \( (1 - x) \)
Now, we need to multiply this expansion by \( (1 - x) \):
\[ (1 - x) \sum_{k=0}^{100} \binom{100}{k} (-1)^k x^{3k} \]
This results in:
\[ \sum_{k=0}^{100} \binom{100}{k} (-1)^k x^{3k} - \sum_{k=0}^{100} \binom{100}{k} (-1)^k x^{3k+1} \]

### Step 5: Identify Coefficient of \( x^{256} \)
To find the coefficient of \( x^{256} \), we need to consider:
1. From the first sum, we need \( 3k = 256 \) which gives \( k = \frac{256}{3} \) (not an integer).
2. From the second sum, we need \( 3k + 1 = 256 \) which gives \( k = \frac{255}{3} = 85 \) (an integer).

Thus, the coefficient we need is:
\[ -\binom{100}{85} (-1)^{85} = \binom{100}{85} \]

### Step 6: Simplify the Coefficient
Using the property of binomial coefficients:
\[ \binom{n}{r} = \binom{n}{n-r} \]
We can write:
\[ \binom{100}{85} = \binom{100}{15} \]

### Final Answer
Thus, the coefficient of \( x^{256} \) in the expansion is:
\[ \boxed{\binom{100}{15}} \]