题目

已知数列{an}的前n项和sn=n^2,设bn=an/3^n,记数列{bn}的前n项和为Tn,求通项公式

提问时间：2020-10-18

答案

a(1) = s(1) = 1^2 = 1,
a(n) = s(n) - s(n-1) = n^2 - (n-1)^2 = 2n-1,n = 1,2,...
b(n) = a(n)/3^n = (2n-1)/3^n = 2*n/3^n - (1/3)^n
设c(n) = n/3^n,n = 1,2,...
G(n) = c(1) + c(2) + ...+ c(n)
= 1/3^1 + 2/3^2 + ...+ n/3^n,
3G(n) = 1 + 2/3 + ...+ n/3^(n-1),
2G(n) = 1 + 1/3 + ...+ 1/3^(n-1) - n/3^n
= [1 - 1/3^n]/[1 - 1/3] - n/3^n
= 3/2[1 - 1/3^n] - n/3^n,
T(n) = b(1) + b(2) + ...+ b(n)
= 2G(n) - [1/3 + (1/3)^2 + ...+ (1/3)^n]
= 3/2[1 - 1/3^n] - n/3^n - 1/3[1 - 1/3^n]/[1-1/3]
= 3/2 - (3/2)(1/3^n) - n/3^n - 1/2 + (1/2)(1/3^n)
= 1 - (1+n)/3^n