Question

关于数列的一道题，求解

Answer 1

(1)
a1+3.a2+3^2.a3+...+3^(n-1).an= n/3 (1)
n=1
a1= 1/3

a1+3.a2+3^2.a3+...+3^(n-2).a(n-1)= (n-1)/3 (2)
(1)-(2)
3^(n-1).an = 1/3
an = (1/3)^n
(2)

let
S = 1.3^1+2.3^2+...+n.3^n (1)
3S = 1.3^2 + 2./3^3+...+n.3^(n+1) (2)
(2)-(1)
2S = n.3^(n+1) -(3+3^2+...+3^n)
= n.3^(n+1) -(3/2)(3^n-1)
S = 2n.3^(n+1) -3(3^n-1)
= 3+(6n-3).3^n
bn = n/an

= n.3^n
Sn =b1+b2+...+bn
= S
=3+(6n-3).3^n