题目

(理)已知数列{an}的各项均为正数,Sn为其前n项和,对于任意n∈N*,满足关系Sn=2an-2.(1)求数列{an}的通项公式;(2)设数列{bn}的前n项和为Tn,且bn=,求证:对任意正整数n,总有Tn＜2;(3)在正数数列{cn}中,设(cn)n+1=an+1(n∈N*),求数列{lncn}中的最大项.(文)已知数列{xn}满足xn+1-xn=()n,n∈N*,且x1=1.设an=xn,且T2n=a1+2a2+3a3+…+ (2n-1)a2n-1+2na2n.(1)求xn的表达式;(2)求T2n;(3)若Qn=1(n∈N*),试比较9T2n与Qn的大小,并说明理由. 答案：(理)(1)解：∵Sn=2an-2(n∈N*),                                               ①∴Sn-1=2an-1-2(n≥2,n∈N*).                                               ②  ①-②,得an=2an-2an-1(n≥2,n∈N*).∵an≠0,∴=2(n≥2,n∈N*),即数列{an}是等比数列.                                                      ∵a1=S1,∴a1=2a1-2,即a1=2.∴an=2n(n∈N*).                                                           (2)证明:∵对任意正整数n,总有bn=,                          ∴Tn==1+1＜2.                                  (3)解:由(cn)n+1=an+1(n∈N*),知lncn=.令f(x)=,则f′(x)=.∵在区间(0,e)上,f′(x)＞0,在区间(e,+∞)上,f′(x)＜0,∴在区间(e,+∞)上f(x)为单调递减函数.                                         ∴n≥2且n∈N*时,{lncn}是递减数列.又lnc1＜lnc2,∴数列{lncn}中的最大项为lnc2=ln3.                              (文)解：(1)∵xn+1-xn=()n,∴xn=x1+(x2-x1)+(x3-x2)+…+(xn-xn-1)=1+()+()2+…+()n-1==.                                                           当n=1时上式也成立,∴xn=(n∈N*).                                                 (2)an=.∵T2n=a1+2a2+3a3+…+(2n-1)a2n-1+2na2n=()2+2()3+3()4+…+(2n-1)()2n+2n()2n+1,                        ①∴T2n=()3+2()4+3()5+…+(2n-1)()2n+1+2n()2n+2.              ②①-②,得T2n=()2+()3+…+()2n+1-2n()2n+2.                        ∴T2n=-2n()2n+2=.∴T2n=.                             (3)由(2)可得9T2n=.又Qn=,当n=1时,22n=4,(2n+1)2=9,∴9T2n＜Qn;                                         当n=2时,22n=16,(2n+1)2=25,∴9T2n＜Qn;                                       当n≥3时,22n=［(1+1)n］2=()2＞(2n+1)2,∴9T2n＞Qn.综上所述,当n=1,2时,9T2n＜Qn;当n≥3时,9T2n＞Qn.

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