Mathematics - Binomial Theorem Question with Solution | TestHub

Let, $\mathrm{y}={\left(1-\mathrm{x}\right)}^{2008}{\left(1+\mathrm{x}+{\mathrm{x}}^2\right)}^{2007}$

$\Rightarrow \mathrm{y}=\left(1-\mathrm{x}\right){\left(1-\mathrm{x}\right)}^{2007}{\left(1+\mathrm{x}+{\mathrm{x}}^2\right)}^{2007}$

$\Rightarrow \mathrm{y}=\left(1-\mathrm{x}\right){\left[\left(1-\mathrm{x}\right)\left(1+\mathrm{x}+{\mathrm{x}}^2\right)\right]}^{2007}$

$\Rightarrow \mathrm{y}=\left(1-\mathrm{x}\right){\left(1-{\mathrm{x}}^3\right)}^{2007}$

$\Rightarrow \mathrm{y}=\left(1-\mathrm{x}\right)\left({}^{2007}\mathrm{C}_0-{}^{2007}\mathrm{C}_1\left({\mathrm{x}}^3\right)+......\right)$

General term in the expansion is given by,

${\mathrm{T}}_{\mathrm{r}+1}=\left(1-\mathrm{x}\right)\left({}^{2007}\mathrm{C}_{\mathrm{r}} {\mathrm{x}}^{3\mathrm{r}} {\left(-1\right)}^{\mathrm{r}}\right)$

$\Rightarrow {\mathrm{T}}_{\mathrm{r}+1}{=}^{2007}{\mathrm{C}}_{\mathrm{r}} {\mathrm{x}}^{3\mathrm{r}} {\left(-1\right)}^{\mathrm{r}}-{ }^{2007}{\mathrm{C}}_{\mathrm{r}} {\mathrm{x}}^{3\mathrm{r}+1} {\left(-1\right)}^{\mathrm{r}}$

For coefficient of ${\mathrm{x}}^{12}$,

$3\mathrm{r}=2012$

But, $\mathrm{r}\ne \frac{2012}{3}$

Also, $3\mathrm{r}+1=2012$

$\Rightarrow 3\mathrm{r}=2011$

But, $\mathrm{r}\ne \frac{2011}{3}$

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

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