Question

已知数列｛an｝满足a1=3,an+1-3an=3^n(n∈N^*),数列｛bn｝满足bn=an/

已知数列｛an｝满足a1=3,an+1-3an=3^n(n∈N^*),数列｛bn｝满足bn=an/3n,(1)证明数列｛bn｝是等比数列并求数列｛bn｝的通项公式，(2)求数列｛an｝的前n项和sn

Answer 1

(1)
a(n+1)-3an=3^n

a(n+1)/3^(n+1)-an/3^n=1/3
{an/3^n}是等差数列,d=1/3
bn-b1= (n-1)/3
bn = (n+2)/3
(2)
an = [(n+2)/3].3^n
= n(1/3)^(n-1) + 2.(1/3)^(n-1)
Sn=a1+a2+...+an
=[∑(i:1->n)i (1/3)^i ] + 3[1-(1/3)^n]
let
S = 1.(1/3)^0+2.(1/3)^1+...+n.(1/3)^(n-1) (1)
(1/3)S = 1.(1/3)^1+2.(1/3)^2+...+n.(1/3)^n (2)
(1)-(2)
(2/3)S = [1+ 1/3+...+1/3^(n-1)] - n(1/3)^n
= (3/2)(1-(1/3)^n) - n(1/3)^n
S =(9/4)(1-(1/3)^n) - (3/2).3^n
Sn = S +3[1-(1/3)^n]
=(9/4)(1-(1/3)^n) - (3/2).3^n + 3[1-(1/3)^n]