已知首项都为1的俩个数列,an,bn,满足anbn+1–an+1bn+2bn+1bn=0,若bn=3^n-1,求数列an的前n项和Sn

Question

已知首项都为1的俩个数列,an,bn,满足anbn+1–an+1bn+2bn+1bn=0,若bn=3^n-1,求数列an的前n项和Sn

Answer 1

anb(n+1)-a(n+1)bn+2b(n+1)bn=0
an.3^n -a(n+1).3^(n-1) +2.3^(2n-1) =0
3an -a(n+1) +2.3^n =0
a(n+1)/3^(n+1) -an/3^n = 2/3
=>{an/3^n} 是等差数列,d=2/3
an/3^n - a1/3 = 2(n-1)/3
an/3^n = (2n-1)/3
an = (2n-1).3^(n-1)
let
S = 1.3^0+2.3^1+.+n.3^(n-1) (1)
3S = 1.3^1+2.3^2+.+n.3^n (2)
(2)-(1)
2S = n.3^n - (1+3+...+3^(n-1))
=n.3^n - (1/2)(3^n -1)
an = (2n-1).3^(n-1)
= 2.[n.3^(n-1)] - 3^(n-1)
Sn =a1+a2+...+an
= 2S -(1/2)(3^n -1)
=n.3^n - (3^n -1)
= 1+(n-1).3^n