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7.Binomial Theorem
hard
$(1-x)^{2008}\left(1+x+x^2\right)^{2007}$ ના વિસ્તરણમાં $x^{2012}$ નો સહગુણક........................છે.
A
$0$
B
$11$
C
$2$
D
$3$
(JEE MAIN-2024)
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 $\mathrm{x}^{2012}$.
So coefficient of $x^{2012}=0$
Standard 11
Mathematics
