Verified Solution • 2025–26 Session moderate

QJEEM24Mafe83: JEE Main 2024 Mathematics - Class 11 JEEM24Mafe83 Solution: The coefficient of in the expansion of is equal to _________.

The coefficient of x2012x^{2012} in the expansion of (1−x)2008(1+x+x2)2007(1-x)^{2008}\left(1+x+x^2\right)^{2007} is equal to _________.

Step-by-Step Verified Solution

amp;(1−x)(1−x)2007(1+x+x2)2007amp;(1−x)(1−x3)2007amp;(1−x)(2007C0−2007C1(x3)+…….)\begin{aligned}& (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)\end{aligned}

General term

amp;(1−x)((−1)r2007Crx3r)amp;(−1)r2007Crx3r−(−1)r2007Crx3r+1amp;3r=2012amp;r≠20123amp;3r+1=2012amp;3r=2011amp;r≠20113\begin{aligned}& (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}\end{aligned}

Hence there is no term containing x2012\mathrm{x}^{2012} .

So coefficient of x2012=0\mathrm{x}^{2012}=0

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