The coefficient of x 2012 in the expansion of (1-x) 2008 (1+x+x 2 ) 2007 is equal to

Mathematics · Binomial Theorem

JEE Main 2024 — 27 January, Shift 2 — Question 25

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

Numerical answer — enter this value.

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

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

$(1-x)\left({ }^{2007} C_{0}-{ }^{2007} C_{1}\left(x^{3}\right)+\ldots \ldots ..\right)$

General term $(1-x)\left((-1)^{r}{ }^{2007} C_{r}{ }^{3 r}\right)$

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

$3 \mathrm{r}=2012$

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

$3 \mathrm{r}+1=2012$

$3 \mathrm{r}=2011$

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

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

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

Answer key and solution verified before publishing.

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