Question

(x-

1

x6

)2009的二項(xiàng)展開式中常數(shù)項(xiàng)為（　�。�

A．C₂₀₀₉²⁸⁶

B．C₂₀₀₉²⁸⁷

C．-C₂₀₀₉²⁸⁶

D．-C₂₀₀₉²⁸⁷

Answer 1

分析：由二項(xiàng)展開式的通項(xiàng)T_r+1=

C

r 2009

•x^2009-r•（-x^-6）^r=（-1）^r•

C

r 2009

•x^2009-7r中x的次數(shù)為0即可得答案．

解答：解：∵T_r+1=

C

r 2009

•x^2009-r•（-x^-6）^r=（-1）^r•

C

r 2009

•x^2009-7r，
∴令2009-7r=0得：r=287．
∴(x-

1

x6

)2009的二項(xiàng)展開式中常數(shù)項(xiàng)為T₂₈₈=（-1）²⁸⁷•

C

287 2009

=-

C

287 2009

．
故選D．

點(diǎn)評(píng)：本題考查二項(xiàng)式定理的應(yīng)用，著重考查二項(xiàng)展開式的通項(xiàng)公式，屬于中檔題．