Question

Coefficient of x²⁰¹² in (1-x)²⁰⁰⁸(1+x+x²)²⁰⁰⁷ is equal to ___

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is A

To determine the coefficient of \(x^{2012}\) in the expression \((1-x)^{2008}(1+x+x^2)^{2007}\), we analyze each factor separately.

For the binomial expansion \((1-x)^{2008}\), the general term is \(\binom{2008}{k}(-1)^kx^k\).

For the trinomial expansion \((1 + x + x^2)^{2007}\), the general term, via the multinomial theorem, is \(\frac{2007!}{a!b!c!}x^{b+2c}\), where \(a + b + c = 2007\).

The product requires the sum of terms where the powers of \(x\) from each factor add up to 2012. Thus, if the first factor contributes \(x^k\), the second factor must contribute \(x^{2012-k}\). This means \(b+2c = 2012-k\).

We have the system of equations:

• \(b + 2c = 2012 - k\)
• \(a + b + c = 2007\)

Subtracting the second equation from the first yields \(c - a = 5 - k\).

The coefficient of \(x^{2012}\) is the sum of \(\binom{2008}{k}(-1)^k \cdot \frac{2007!}{a!b!c!}\) over all valid integer, non-negative values of \(a, b, c,\) and \(k\) satisfying the derived relationships. However, examination of these conditions reveals no combination of \(a, b, c,\) and \(k\) that can produce \(x^{2012}\) from the given expression. Therefore, the coefficient is 0.