Question

已知a<b,且a^2-a-6=0,b^2-b-6=0,数列{an}{bn}满足a1=1,a2=-6a,a(n+1)=6an-9a(n-1)(n>=2),bn=a(n+1)-ban

(1)求证数列{bn}是等比数列
（2）已知数列{cn}满足cn=an/3^n,试建立数列{cn}的递推公式（要求不含an或bn）
(3)若数列{an}的前n项和为Sn，求Sn

Answer 1

(1)

(2)

(3)

Answer 2

(1)
a^2-a-6=0
(a-3)(a+2)=0
a=3 or -2
b^2-b-6=0

b=3 or -2
a<b
=> a=-2 , b=3
a1=1, a2=-6a= 12

a(n+1)=6an-9a(n-1)
a(n+1)- 3an = 3(an-3a(n-1))
[a(n+1)- 3an]/(an-3a(n-1))=3
(an-3a(n-1))/(a2-3a1) = 3^(n-2)
an-3a(n-1) =3^n

an/3^n -a(n-1)/3^(n-1) =1
an/3^n - a1/3 = n-1
an/3^n=(n-1) + 1/3
= (1/3)(3n- 2)
an = 3^(n-1) .(3n-2)

bn=a(n+1)-ban
= 3^(n) .(3n+1) - 3^(n) .(3n-2)
= 3^(n+1)
{bn}是等比数列
cn = an/3^n
= (1/3)(3n-2)

an = 3^(n-1) .(3n-2)
= 3(n.3^(n-1)) - 2(3^(n-1))
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=3
1.3^0+2.3^1+...+n.3^(n-1) = (1/4) { n3^(n+1)- (n+1)3^n +1 }

Sn = a1+a2+a3+..+an
= (3/4) { n3^(n+1)- (n+1)3^n +1 } - (3^n-1)
= (3/4)3^n {2n-5} +7/4