Question

在(1+x-x²)⁶的展开式中x⁵的系数为(　　)

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

Answer 1

思路解析：设(1+x-x²)⁶=［1+(x-x²)］⁶的展开式的通项为T_r+1,

则T_r+1=(x-x²)^r(r=0,1,2,…,6).二项式(x-x²)^r展开式的通项为

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

(1+x-x²)⁶的展开式的通项公式为T_r+1=

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

(1+x-x²)⁶展开式中含x⁵项的系数为(-1)²+(-1)+(-1)⁰=6.

答案：C