若(2x−1x)n展开式中含1x2项的系数与含1x4项的系数之比为-5,则n等于(  ) A.4 B.6 C.8 D.10

问题描述:

(2x−

1
x
)n展开式中含
1
x2
项的系数与含
1
x4
项的系数之比为-5,则n等于(  )
A. 4
B. 6
C. 8
D. 10

(2x−1x)n展开式的通项为Tr+1=Crn(2x)n−r(−1x)r=(-1)r2n-rCnrxn-2r令n-2r=-2得r=n+22故含1x2的系数为(−1)n+222n−22Cn+22n令n-2r=-4得r=n+42故含1x4项的系数为(−1)n+422n−42Cn+42n∴(−1)n+222n−22Cn+22n(...