Coefficient of `x^(2012)` in `(1-x)^(2008)(1+x+x^2)^2007`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the coefficient of \( x^{2012} \) in the expression \( (1-x)^{2008}(1+x+x^2)^{2007} \), we can follow these steps:

### Step 1: Simplify the Expression
We start with the expression:
\[ (1-x)^{2008}(1+x+x^2)^{2007} \]
We can rewrite \( (1+x+x^2)^{2007} \) using the identity:
\[ 1+x+x^2 = \frac{1-x^3}{1-x} \]
Thus, we have:
\[ (1+x+x^2)^{2007} = \left(\frac{1-x^3}{1-x}\right)^{2007} = (1-x^3)^{2007}(1-x)^{-2007} \]
Now substituting this back into our original expression gives:
\[ (1-x)^{2008}(1-x^3)^{2007}(1-x)^{-2007} = (1-x)^{2008 - 2007}(1-x^3)^{2007} = (1-x)(1-x^3)^{2007} \]

### Step 2: Expand \( (1-x)(1-x^3)^{2007} \)
Now we need to expand \( (1-x)(1-x^3)^{2007} \):
\[ (1-x)(1-x^3)^{2007} = (1-x) \sum_{r=0}^{2007} \binom{2007}{r} (-1)^r x^{3r} \]
This gives us:
\[ = \sum_{r=0}^{2007} \binom{2007}{r} (-1)^r x^{3r} - x \sum_{r=0}^{2007} \binom{2007}{r} (-1)^r x^{3r} \]
The first term contributes \( x^{3r} \) and the second term contributes \( x^{3r+1} \).

### Step 3: Find Coefficient of \( x^{2012} \)
We need to find the coefficient of \( x^{2012} \) in the expanded expression:
1. From the first sum \( \sum_{r=0}^{2007} \binom{2007}{r} (-1)^r x^{3r} \):
- We set \( 3r = 2012 \) which gives \( r = \frac{2012}{3} \). This is not an integer, so there is no contribution from this term.

2. From the second sum \( -x \sum_{r=0}^{2007} \binom{2007}{r} (-1)^r x^{3r} \):
- We set \( 3r + 1 = 2012 \) which gives \( 3r = 2011 \) or \( r = \frac{2011}{3} \). This is also not an integer, so there is no contribution from this term either.

### Conclusion
Since both possible contributions to \( x^{2012} \) yield no integer values for \( r \), we conclude that the coefficient of \( x^{2012} \) in the original expression is:
\[ \boxed{0} \]