Question

(理)已知數列{a_n}的各項均為正數,S_n為其前n項和,對于任意n∈N^*,滿足關系S_n=2a_n-2.

(1)求數列{a_n}的通項公式;

(2)設數列{b_n}的前n項和為T_n,且b_n=,求證:對任意正整數n,總有T_n＜2;

(3)在正數數列{c_n}中,設(c_n)ⁿ⁺¹=a_n+1(n∈N^*),求數列{lnc_n}中的最大項.

(文)已知數列{x_n}滿足x_n+1-x_n=()ⁿ,n∈N^*,且x₁=1.設a_n=x_n,且T_2n=a₁+2a₂+3a₃+…+ (2n-1)a_2n-1+2na_2n.

(1)求x_n的表達式;

(2)求T_2n;

(3)若Q_n=1(n∈N^*),試比較9T_2n與Q_n的大小,并說明理由.

Answer 1

(理)(1)解：∵S_n=2a_n-2(n∈N^*),                                               ①

∴S_n-1=2a_n-1-2(n≥2,n∈N^*).                                               ② 

①-②,得a_n=2a_n-2a_n-1(n≥2,n∈N^*).

∵a_n≠0,∴=2(n≥2,n∈N^*),

即數列{a_n}是等比數列.                                                      

∵a₁=S₁,

∴a₁=2a₁-2,即a₁=2.

∴a_n=2ⁿ(n∈N^*).                                                          

(2)證明:∵對任意正整數n,總有b_n=,                         

∴T_n=

=1+1＜2.                                 

(3)解:由(c_n)ⁿ⁺¹=a_n+1(n∈N^*),知lnc_n=.

令f(x)=,則f′(x)=.

∵在區間(0,e)上,f′(x)＞0,在區間(e,+∞)上,f′(x)＜0,

∴在區間(e,+∞)上f(x)為單調遞減函數.                                         

∴n≥2且n∈N^*時,{lnc_n}是遞減數列.

又lnc₁＜lnc₂,∴數列{lnc_n}中的最大項為lnc₂=ln3.                             

(文)解：(1)∵x_n+1-x_n=()ⁿ,

∴x_n=x₁+(x₂-x₁)+(x₃-x₂)+…+(x_n-x_n-1)

=1+()+()²+…+()^n-1

=

=.                                                          

當n=1時上式也成立,

∴x_n=(n∈N^*).                                                

(2)a_n=.

∵T_2n=a₁+2a₂+3a₃+…+(2n-1)a_2n-1+2na_2n

=()²+2()³+3()⁴+…+(2n-1)()²ⁿ+2n()²ⁿ⁺¹,                        ①

∴T_2n=()³+2()⁴+3()⁵+…+(2n-1)()²ⁿ⁺¹+2n()²ⁿ⁺².              ②

①-②,得T_2n=()²+()³+…+()²ⁿ⁺¹-2n()²ⁿ⁺².                       

∴T_2n=-2n()²ⁿ⁺²

=.

∴T_2n=.                             

(3)由(2)可得9T_2n=.

又Q_n=,

當n=1時,2²ⁿ=4,(2n+1)²=9,∴9T_2n＜Q_n;                                        

當n=2時,2²ⁿ=16,(2n+1)²=25,∴9T_2n＜Q_n;                                      

當n≥3時,2²ⁿ=［(1+1)ⁿ］²=()²＞(2n+1)²,

∴9T_2n＞Q_n.

綜上所述,當n=1,2時,9T_2n＜Q_n;當n≥3時,9T_2n＞Q_n.