Binomial Theorem Question 5

Question 5 - 2024 (27 Jan Shift 2)

The coefficient of $x^{2012}$ in the expansion of $(1-x)^{2008}\left(1+x+x^{2}\right)^{2007}$ is equal to

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Solution

$(1-x)(1-x)^{2007}\left(1+x+x^{2}\right)^{2007}$

$(1-x)\left(1-x^{3}\right)^{2007}$

$(1-x)\left({ }^{2007} C _0-{ }^{2007} C _1\left(x^{3}\right)+\ldots \ldots\right)$

General term

$(1-x)\left((-1)^{r 2007} C _r x^{3 r}\right)$

$(-1)^{r 2007} C _r x^{3 r}-(-1)^{r 2007} C _r x^{3 r+1}$

$3 r=2012$

$r \neq \frac{2012}{3}$

$3 r+1=2012$

$3 r=2011$

$r \neq \frac{2011}{3}$

Hence there is no term containing $x^{2012}$.

So coefficient of $x^{2012}=0$