Question

(理)已知数列{a_n}满足a_n=2a_n-1+2ⁿ-1(n≥2),a₁=5,b_n=.

(1)证明{b_n}为等差数列;

(2)求数列{a_n}的前n项和S_n.

(文)已知数列{a_n}的前n项和为S_n,且当n∈N^*时满足S_n=-3n²+6n,数列{b_n}满足b_n=()^n-1,数列{c_n}满足c_n=a_n·b_n.

(1)求数列{a_n}的通项公式a_n;

(2)求数列{c_n}的前n项和T_n.

Answer 1

(理)(1)证明:∵b_n=

=+1=b_n-1+1(n≥2),                                                     

∴b_n-b_n-1=1(n≥2).

∴{b_n}是公差为1,首项为b₁==2的等差数列.                              

(2)解：由(1)知b_n=2+(n-1)·1=n+1,

即=n+1,∴a_n=(n+1)2ⁿ+1.                                               

∴S_n=［2·2+3·2²+4·2³+…+(n+1)2ⁿ］+n.

令T_n=2·2+3·2²+…+n·2^n-1+(n+1)·2ⁿ,

∴2T_n=2·2²+…+n·2ⁿ+(n+1)2ⁿ⁺¹.

∴-T_n=2·2+1·2²+…+1·2ⁿ-(n+1)2ⁿ⁺¹

=4+-(n+1)2ⁿ⁺¹

=4+2ⁿ⁺¹-4-n·2ⁿ⁺¹-2ⁿ⁺¹=-n·2ⁿ⁺¹.

∴T_n=n·2ⁿ⁺¹.                                                             

∴S_n=n·2ⁿ⁺¹+n.                                                           

(文)解：(1)当n=1时,a₁=S₁=3,                                             

当n≥2时,a_n=S_n-S_n-1=9-6n,                                                  

∴a_n=9-6n.                                                              

(2)∵b_n=()^n-1,c_n=a_n·b_n=·()^n-1=(3-2n)·()ⁿ,

∴T_n=-()²-3·()³-…-(2n-3)·()ⁿ,                                        

T_n=()²-()³-…-(2n-5)·()ⁿ-(2n-3)()ⁿ⁺¹.

∴T_n=-2·()²-2·()³-…-?2·?()ⁿ+(2n-3)·()ⁿ⁺¹

=-2·+(2n-3)·()ⁿ⁺¹,                                        

∴T_n=(2n+1)()ⁿ-1.