Question

若(2x-

1

x

)ⁿ展開(kāi)式中含

1

x2

項(xiàng)的系數(shù)與含

1

x4

項(xiàng)的系數(shù)之比為-5，則n等于（　�。�

A．4

B．6

C．8

D．10

Answer 1

(2x-

1

x

)n展開(kāi)式的通項(xiàng)為
Tr+1=

C

rn

(2x)n-r(-

1

x

)r=（-1）^r2^n-rC_n^rx^n-2r
令n-2r=-2得r=

n+2

2

故含

1

x2

的系數(shù)為(-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

項(xiàng)的系數(shù)為(-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
解得n=6
故選B