定义:若数列{an}对任意n∈N*,满足a(n+2)-a(n+1)/a(n+1)-an=k(k为常数)称数列{an}为等差比数列.(1)若数列{an}前n项和Sn=3(an-2),qiu {an}的通项公式,并判断该数列是否为等差比数列；(2)若数列{an}为等差数列,

定义:若数列{an}对任意n∈N*,满足a(n+2)-a(n+1)/a(n+1)-an=k(k为常数)称数列{an}为等差比数列.(1)若数列{an}前n项和Sn=3(an-2),qiu {an}的通项公式,并判断该数列是否为等差比数列；(2)若数列{an}为等差数列,
定义:若数列{an}对任意n∈N*,满足a(n+2)-a(n+1)/a(n+1)-an=k(k为常数)称数列{an}为等差比数列.
(1)若数列{an}前n项和Sn=3(an-2),qiu {an}的通项公式,并判断该数列是否为等差比数列；
(2)若数列{an}为等差数列,是判断{an}是否一定为等差比数列,并说明理由；
(3)若数列{an}为等差比数列,定义中常数k=2,a2=3,a1=1,数列{（2n-1)/(an+1)}的前n项和为Tn,求证：Tn＜3.

定义:若数列{an}对任意n∈N*,满足a(n+2)-a(n+1)/a(n+1)-an=k(k为常数)称数列{an}为等差比数列.(1)若数列{an}前n项和Sn=3(an-2),qiu {an}的通项公式,并判断该数列是否为等差比数列；(2)若数列{an}为等差数列,
(1)Sn=3(an-2)
S(n-1)=3(a(n-1)-2)
两式相减,得
an=3an-3a(n-1)
即an=3/2a(n-1)
所以{an}是等比数列,公比为3/2
而a1=S1=3(a1-2),得a1=3
an=3*(3/2)^(n-1)
(a(n+2)-a(n+1))/(a(n+1)-an)=[(3/2)^2-3/2]/(3/2-1)=3/2
所以该数列是等差比数列
(2)不一定,如果公差为0,a(n+1)-an=0
(3)a(n+2)-a(n+1)=2(a(n+1)-an)
{a(n+1)-an}是等比数列,公比为2
a2-a1=2
所以a(n+1)-an=2^n
an=a1+(a2-a1)+(a3-a2)+...+(an-a(n-1))=1+2+2^2+...+2^(n-1)=2^n-1
所以(2n-1)/(an+1)=(2n-1)/2^n
对于Tn,使用错位相减法
Tn=1/2+3/2^2+3/2^3+...+(2n-3)/2^(n-1)+(2n-1)/2^n①
1/2Tn=1/2^2+3/2^3+5/2^4+...+(2n-3)/2^n+(2n-1)/2^(n+1)②
①-②,得
1/2Tn=1/2+2/2^2+2/2^3+...+2/2^n-(2n-1)/2^(n+1)
=1/2+1/2(1-1/2^(n-1))/(1-1/2)-(2n-1)/2^(n+1)
=3/2-(2n+3)/2^(n+1)
所以Tn=3-(2n+3)/2^n

(1)
Sn=3(an-2)
n=1
a1=3
Sn=3(an-2) (1)
S(n-1) =3(a(n-1)-2) (2)
(1)-(2)
an = 3an-3a(n-1)
an/a(n-1) = 3/2
an/a1 = (...

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(1)
Sn=3(an-2)
n=1
a1=3
Sn=3(an-2) (1)
S(n-1) =3(a(n-1)-2) (2)
(1)-(2)
an = 3an-3a(n-1)
an/a(n-1) = 3/2
an/a1 = (3/2)^(n-1)
an= 2.(3/2)^n
[a(n+2) -a(n+1)]/[a(n+1)-an]
=2(3/2)^(n+1) . (1/2)/ [ 2(3/2)^(n) . (1/2)]
= 3/2
{an}为等差比数列
(2)
{an}为等差数列
=> a(n+2)- a(n+1) = d and a(n+1)- an =d
[a(n+2) -a(n+1)]/[a(n+1)-an] =d/d =1
=>{an}为等差比数列
(3)
{an}为等差比数列
[a(n+2) -a(n+1)]/[a(n+1)-an] = 2
[a(n+2) -a(n+1)]/[a2-a1] = 2^n
a(n+2) -a(n+1) = 2^(n+1)
an -a(n-1) =2^(n-1)
an -a1 = 2+2^2+...+2^(n-1)
an = 2^0+2^1+...+2^(n-1)
= 2^n -1
bn =(2n-1)/(an+1)
= (2n-1) .(1/2)^n
= n.(1/2)^(n-1) - (1/2)^n
Tn =b1+b2+...+bn
=[ ∑(i:1->n) i(1/2)^(i-1) ] - [1-(1/2)^n]
consider
1+x+x^2+...+x^n = [x^(n+1) -1]/(x-1)
1+2x+...+nx^(n-1)=[ [x^(n+1) -1]/(x-1) ]'
= [nx^(n+1) -(n+1)x^n + 1] /(x-1)^2
put x=1/2
∑(i:1->n) i(1/2)^(i-1)
=4[n.(1/2)^(n+1) -(n+1).(1/2)^n + 1]
=4[1- (n+2).(1/2)^(n+1)]
Tn =[ ∑(i:1->n) i(1/2)^(i-1) ] - [1-(1/2)^n]
= 4[1- (n+2).(1/2)^(n+1)] - [1-(1/2)^n ]
= 3 - (2n+3).(1/2)^n

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