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−

1
x
)n展开式的通项为
Tr+1
C rn
(2x)n−r(−
1
x
)
r
=(-1)r2n-rCnrxn-2r
令n-2r=-2得r=
n+2
2

故含
1
x2
的系数为(−1)
n+2
2
2
n−2
2
C
n+2
2
n

令n-2r=-4得r=
n+4
2

故含
1
x4
项的系数为(−1)
n+4
2
2
n−4
2
C
n+4
2
n

(−1)
n+2
2
2
n−2
2
C
n+2
2
n
(−1)
n+4
2
2
n−4
2
C
n+4
2
n
=−5

将n=4,6,8,10代入检验得n=6
故选B