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Q.

The coefficient of  x2012 in the expansion of (1x)2008(1+x+x2)2007  is equal to k then k2=  ______

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Detailed Solution

 (1x)(1x)2007(1+x+x2)2007    (1x)(1x3)2007   (1x)(2007C02007C1(x3)+......)
 General term
 (1x)((1)r2007Crx3r)    (1)Cr   r2007x3r(1)Cr   r2007x3r+1 3r=2012   r20123   3r+1=2012 3r=2011  r20113
 Hence there is no term containing x2012 .
So coefficient of  x2012=0
 

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