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 find the coefficient of \(x^{2012}\) in the given expression \((1-x)^{2008}(1+x+x^2)^{2007}\), we need to examine both components of the multiplication. 

Firstly, consider the binomial expansion of \((1-x)^{2008}\):

• The general term in this expansion is \(\binom{2008}{k}(-x)^k = \binom{2008}{k}(-1)^kx^k\).

Now consider the trinomial expansion of \((1 + x + x^2)^{2007}\):

• The general term in this expansion can be expressed using the multinomial theorem as: \(\frac{2007!}{a!b!c!} \cdot 1^a \cdot x^b \cdot (x^2)^c = \frac{2007!}{a!b!c!}x^{b+2c}\), where \(a + b + c = 2007\).

We need the coefficient of \(x^{2012}\) in the product:

From the above expansions, to have \(x^{2012}\):

• The term \((1-x)^{2008}\) contributes \(x^k\).
• The term \((1+x+x^2)^{2007}\) will contribute \(x^{2012-k}\).

Since \(b + 2c = 2012 - k\) and \(a + b + c = 2007\), we have:

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

Subtracting the second equation from the first gives:

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

which simplifies to:

\(c - a = 5 - k\)

The coefficient term resulting from \(x^{2012}\) is:

\(\sum \left[ \binom{2008}{k}(-1)^k \cdot \frac{2007!}{a!b!c!} \right]\), where these terms satisfy the above equations.

Each value must logically satisfy the conditions of integer and non-negative requirements for \(a, b, c\), not possible due to constraints from the multinomial equation and range.

After examining possible integer solutions, no such combination of \(a, b, c\) and \(x^{2012}\) in the original expression can be concluded mathematically to be \(0\).

Answer 2

Consider the expansion:
\((1 - x)^{2008} \quad \text{and} \quad (1 + x + x^2)^{2007}.\)

The expansion of \((1 - x)^{2008}\) yields terms of the form \((-1)^k \binom{2008}{k} x^k\) for \(k \geq 0\). Thus, it contains only terms with non-positive powers of \(x\) (i.e., \(x^0, x^1, x^2, \dots\)).

The expansion of \((1 + x + x^2)^{2007}\) contains terms of the form \(x^m\), where \(m\) is a non-negative integer ranging from 0 to 4014 (since the highest power in the expansion occurs when all factors contribute \(x^2\)).

To find the coefficient of \(x^{2012}\) in the product:  
\((1 - x)^{2008} \cdot (1 + x + x^2)^{2007},\)
we note that there is no term in \((1 - x)^{2008}\) with a negative power of \(x\) to combine with terms in \((1 + x + x^2)^{2007}\) such that the resulting power of \(x\) is 2012. Therefore, the coefficient of \(x^{2012}\) in the expansion is:  0

The correct option is (A) : 0