求解高中数列题目

Question

如题所述

Answer 1

(1)
Sn = 3^n+k
n=1, a1= 3+k
an = Sn -S(n-1)
=2. 3^(n-1)
a1= 2 = 3+k
k = -1
(2)
a(n+1)/2 = (4+k)^(an.bn)
3^n= 3^[2.3^(n-1) .bn]
n = 2.3^(n-1) .bn
bn = (1/2) [n.(1/3)^(n-1)]
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 = (3/2)[(3/2)( 1- 1/3^n) - n(1/3)^n]

Tn =b1+b2+...+bn
=(1/2)S
=(3/4)[(3/2)( 1- 1/3^n) - n(1/3)^n]