设数列{an}的前n项和是Sn,Sn满足：S1=a(a≠0),Sn+1 =3Sn+1(n∈N*),数列{bn}满足：bn

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设数列{an}的前n项和是Sn,Sn满足：S1=a(a≠0),Sn+1 =3Sn+1(n∈N*),数列{bn}满足：bn+2 =2bn+1-bn,且
设数列{an}的前n项和是Sn,Sn满足：S1=a(a≠0),Sn+1 =3 Sn+1(n∈N*),数列{bn}满足：bn+2 =2bn+1-bn (n∈N*),且b3=3,b5=9.
(1)分别求数列{an}和{bn}的通项公式；
(2)若a=1,若对任意的n∈N*,都有(Sn+1/2)*k≥bn恒成立,求实数k的取值范围.

(1)
S(n+1)=3Sn+1
S(n+1)+1/2=3(Sn+1/2)
[S(n+1)+1/2]/(Sn+1/2)=3
(Sn+1/2)/(S1+1/2)=3^(n-1)
Sn = -1/2 + (a+1/2)3^(n-1)
an = Sn-S(n-1)
= (2a+1).3^(n-2)
ie
an = a ; n=1
= (2a+1).3^(n-2) ; n>=2
b(n+2) = 2b(n+1)-bn
The aux. equation
x^2-2x+1 =0
x=1
let
bn = A+Bn
b3= A+3B =3 (1)
b5= A+5B = 9 (2)
(2)-(1)
B=3
A=-6
bn =9n-6
(2)
a=1
an =3^(n-1)
Sn= (3^n-1)/2
(Sn+1/2)*k≥bn
(3^n)*k≥ 9n-6
k≥ (9n-6)/3^n
let
f(x)= (9x-6) . 3^(-x)
f'(x) = 3^(-x) [ -(9x-6)ln3 + 9 ]
f'(x) =0
-(9x-6)ln3 + 9 =0
x= (9-6ln3)/(9ln3) =0.243 (max)
max (9n-6)/3^n at n=1
max (9n-6)/3^n = 3/3 =1
ie k≥1

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