Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{2012}$ in the expansion of $(1-x{)}^{2008}(1+x+x^2{)}^{2007}$ is equal to

A Journey Through Binomial Expansion

When you first look at a problem like finding the coefficient of $x^{2012}$ in the expansion of $(1-x{)}^{2008}(1+x+x^2{)}^{2007}$, it is natural to feel a surge of intimidation. The exponents are massive, the expression looks unwieldy, and the prospect of expanding these terms seems like a task for a supercomputer, not a student.

But here is the secret of the JEE Advanced: Complexity is often just a mask for elegance.

The Strategic Split

Our first step is to stop looking at the numbers as obstacles and start seeing them as opportunities. We have $(1-x{)}^{2008}$ and $(1+x+x^2{)}^{2007}$. Notice that $2008$ and $2007$ are almost identical.

In the world of algebra, when you see exponents that are off by one, you should immediately think about 'borrowing.' We can rewrite $(1-x{)}^{2008}$ as $(1-x)\cdot (1-x{)}^{2007}$.

Now, our expression looks like this:

By doing this, we have successfully isolated a single $(1-x)$ term and created a pair of terms with the same exponent, $2007$. This is the moment the problem begins to yield.

The Hidden Identity

Now, we use the power rule $a^n\cdot b^n=(ab{)}^n$. We group the terms with the exponent $2007$ together:

Look closely at the expression inside the bracket: $(1-x)(1+x+x^2)$. If you have been practicing your algebraic identities, this should trigger a recognition.

This is the classic difference of cubes identity: $(a-b)(a^2+ab+b^2)=a^3-b^3$. With $a=1$ and $b=x$, this entire bracket collapses into $1-x^3$. Our massive, intimidating expression has now simplified to:

The Systematic Search

We are now in a much stronger position. We need to find the coefficient of $x^{2012}$ in $(1-x)(1-x^3{)}^{2007}$. Let's expand $(1-x^3{)}^{2007}$ using the general term formula for binomial expansion, $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$.

Here, $n=2007$, $a=1$, and $b=-x^3$:

Now, we multiply this entire series by the $(1-x)$ we left outside:

The Final Verdict

We are hunting for $x^{2012}$. This means we need to check if either $3r=2012$ or $3r+1=2012$ yields an integer value for $r$.

1. For the first part, $3r=2012⟹r=\frac{2012}{3}=\mathrm{670.66...}$ (Not an integer).

2. For the second part, $3r+1=2012⟹3r=2011⟹r=\frac{2011}{3}=\mathrm{670.33...}$ (Not an integer).

Since neither case produces an integer $r$, it means the term $x^{2012}$ simply does not exist in this expansion. It is not there! And if a term is missing, its coefficient is, by definition, 0.

This problem is a masterclass in not panicking. By using the 'borrowing' technique and recognizing the difference of cubes, we turned a mountain into a molehill.

The final answer, 0, is not a failure—it is the result of a rigorous, logical journey. Keep practicing these manipulations; the more you do, the more you will see the beauty hidden behind the numbers.