JEE Mathematics 2024 Question with Solution

Given: We need the coefficient of $x^{2012}$ in $(1-x)^{2008}(1+x+x^2)^{2007}$.

Find: The required coefficient and hence the correct option.

Using the identity

$$1+x+x^2=\frac{1-x^3}{1-x}$$

we get

$$(1-x)^{2008}(1+x+x^2)^{2007}=(1-x)^{2008}\cdot \frac{(1-x^3)^{2007}}{(1-x)^{2007}}=(1-x)(1-x^3)^{2007}$$

Now expand:

$$(1-x^3)^{2007}=\sum_{r=0}^{2007} \binom{2007}{r}(-1)^r x^{3r}$$

So the whole expression becomes

$$(1-x)(1-x^3)^{2007}=(1-x)\sum_{r=0}^{2007} \binom{2007}{r}(-1)^r x^{3r}$$

Hence every term is of the form $x^{3r}$ or $x^{3r+1}$.

But $2012 \equiv 2 \pmod{3}$, so $2012$ is neither of the form $3r$ nor of the form $3r+1$.

Therefore, the coefficient of $x^{2012}$ is $0$.

The correct option is A.

Given: We seek the coefficient of $x^{2012}$ in $(1-x)^{2008}(1+x+x^2)^{2007}$.

Find: Whether any valid combination of powers produces $x^{2012}$.

From

$$(1-x)^{2008}=\sum_{k=0}^{2008} \binom{2008}{k}(-1)^k x^k$$

a general term contributes $x^k$.

In $(1+x+x^2)^{2007}$, a general term is obtained by choosing $1, x, x^2$ respectively $a,b,c$ times, where

$$a+b+c=2007$$

Then the power of $x$ contributed is

$$b+2c$$

So in the product, to obtain $x^{2012}$, we need

$$k+b+2c=2012$$

with

$$a+b+c=2007$$

Subtracting gives

$$k+c-a=5$$

This route is cumbersome. A better observation is that

$$1+x+x^2=\frac{1-x^3}{1-x}$$

which simplifies the entire expression to

$$(1-x)(1-x^3)^{2007}$$

Now $(1-x^3)^{2007}$ contains only powers divisible by $3$, and multiplying by $1-x$ produces only powers of the form $3r$ and $3r+1$.

Since $2012 = 3\cdot 670 + 2$, it is of the form $3r+2$, so no such term appears.

Therefore, the coefficient is $0$, so the correct option is A.

Note: The solution text states the answer correctly as A, although one of its explanations is not rigorous; the simplification above confirms the result cleanly.