Question

在(1+x-x²)⁶的展開式中x⁵的系數(shù)為(　　)

A.4                              B.5                       C.6                       D.7

Answer 1

思路解析：設(shè)(1+x-x²)⁶=［1+(x-x²)］⁶的展開式的通項(xiàng)為T_r+1,

則T_r+1=(x-x²)^r(r=0,1,2,…,6).二項(xiàng)式(x-x²)^r展開式的通項(xiàng)為

t_n+1=(-1)ⁿx^r-n(x²)ⁿ=(-1)ⁿx^r+n(n=0,1,2,…,r).

(1+x-x²)⁶的展開式的通項(xiàng)公式為T_r+1=

令r+n=5,則n=5-r≥0,0≤r≤6,0≤n≤r,r=3,4,5,n=2,1,0.

(1+x-x²)⁶展開式中含x⁵項(xiàng)的系數(shù)為(-1)²+(-1)+(-1)⁰=6.

答案：C