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
$(1-x)\left((-1)^{\mathrm{r} 2007} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{3 \mathrm{r}}\right)$
$(-1)^{\mathrm{r} 2007} \mathrm{C}{\mathrm{r}} \mathrm{x}^{3 \mathrm{r}}-(-1)^{\mathrm{r} 2007} \mathrm{C}{\mathrm{r}} \mathrm{x}^{3 \mathrm{r}+1}$
$3 \mathrm{r}=2012$
$\mathrm{r} \neq \frac{2012}{3}$
$3 \mathrm{r}+1=2012$
$3 \mathrm{r}=2011$
$\mathrm{r} \neq \frac{2011}{3}$
Hence there is no term containing $\mathrm{x}^{2012}$.
So coefficient of $\mathrm{x}^{2012}=0$
