(2)设展开式中第k+1项的系数最大,

　　则由Tk+1=C_5^k (x^(2/3))5-k(3x2)k=3kC_5^k x^((10+4k)/3),

　　得{■(3^k C_5^k≥3^(k"-" 1) C_5^(k"-" 1) "," @3^k C_5^k≥3^(k+1) C_5^(k+1) "," )┤

　　∴7/2≤k≤9/2,

　　∴k=4,

　　即展开式中系数最大的项为T5=C_5^4 (x^(2/3))(3x2)4=405x^(26/3).

9.求证:3n>(n+2)·〖2^n〗^("-" 1)(n∈N+,n>2).

证明因为n∈N+,且n>2,

　　所以3n=(2+1)n展开后至少有4项.

　　(2+1)n=2n+C_n^1·2n-1+...+C_n^(n"-" 1)·2+1≥2n+n·2n-1+2n+1>2n+n·2n-1=(n+2)·2n-1,

　　故3n>(n+2)·2n-1(n∈N+,n>2).

10.求证:1+2+22+...+〖2^5〗^(n"-" 1)(n∈N+)能被31整除.

证明∵1+2+22+...+〖2^5〗^(n"-" 1)=(2^5n "-" 1)/(2"-" 1)

　　=〖2^5〗^n-1=32n-1

　　=(31+1)n-1

　　=C_n^0·31n+C_n^1·31n-1+...+C_n^(n"-" 1)·31+C_n^n-1

　　=31(C_n^0·31n-1+C_n^1·31n-2+...+C_n^(n"-" 1)),

　　显然C_n^0·31n-1+C_n^1·31n-2+...+C_n^(n"-" 1)为整数,

　　∴原式能被31整除.

B组

1.若(x+y)9按x的降幂排列的展开式中,第二项不大于第三项,且x+y=1,xy<0,则x的取值范围是(　　)

A.("-∞," 1/5) B.[4/5 "," +"∞" )

C.("-∞,-" 4/5] D.(1,+∞)

解析:二项式(x+y)9的展开式的通项是

　　Tr+1=C_9^r·x9-r·yr.

依题意,有{■(C_9^1 "·" x^(9"-" 1) "·" y≤C_9^2 "·" x^(9"-" 2) "·" y^2 "," @x+y=1"," @xy<0"," )┤