The coefficient of x2012 in the expansion of (1−x)2008(1+x+x2)2007 is equal to \_\_\_\_\_\_\_\_\_.
Answer: 0
Solutions · 1
Official solution
(1−x)(1−x)2007(1+x+x2)2007(1−x)(1−x3)2007(1−x)(2007C0−2007C1(x3)+…….)
General term
(1−x)((−1)r2007Crx3r)(−1)r2007Crx3r−(−1)r2007Crx3r+13r=2012r=320123r+1=20123r=2011r=32011
Hence there is no term containing x2012.
So coefficient of x2012=0