Dao Trong Luan Coordinator
03/12/2017 at 10:09-
Alchemy 03/12/2017 at 10:53We have: \(2006=2005+1=x+1\)
Hence \(B=x^{2005}-2006x^{2004}+2006x^{2003}-...+2006x+1\)
\(=x^{2005}-\left(x+1\right)x^{2004}+\left(x+1\right)x^{2003}-...+\left(x+1\right)x+1\)
\(=x^{2005}-x^{2005}-x^{2004}+x^{2014}+x^{2013}-...+x^2+x+1\)
\(=x+1=2005+1=2006\)
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HỦY DIỆT THE WORLD 03/12/2017 at 21:42We have: 2006=2005+1=x+12006=2005+1=x+1
Hence B=x2005−2006x2004+2006x2003−...+2006x+1B=x2005−2006x2004+2006x2003−...+2006x+1
=x2005−(x+1)x2004+(x+1)x2003−...+(x+1)x+1=x2005−(x+1)x2004+(x+1)x2003−...+(x+1)x+1
=x2005−x2005−x2004+x2014+x2013−...+x2+x+1=x2005−x2005−x2004+x2014+x2013−...+x2+x+1
=x+1=2005+1=2006