求解:已知An=n/3*n,求数列An的前n项和Sn。

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解决时间 2021-12-03 12:02

提问者网友：疯孩纸
2021-12-03 05:55

最佳答案

五星知识达人网友：一秋
2021-12-03 06:35

let
S = 1.(1/3)^1 + 2.(1/3)^2+....+n.(1/3)^n (1)
(1/3)S = 1.(1/3)^2 + 2.(1/3)^3+....+n.(1/3)^(n+1) (2)
(1)-(2)
(2/3)S = ( 1/3 + 1/3^2+...+1/3^n) -n.(1/3)^(n+1)
= (1/2) ( 1- (1/3)^n ) - n.(1/3)^(n+1)
S = (3/4)( 1- (1/3)^n ) - (1/2)n. (1/3)^n
= 3/4 - (1/4)(2n + 3) .(1/3)^n
Sn =a1+a2+...+an= S = 3/4 - (1/4)(2n + 3) .(1/3)^n追问。。老铁。我怎么看不懂啊。答案是:3/4-(2n+3)/(4×3*n)我看不懂过程才问的。追答答案是:3/4-(2n+3)/(4×3*n)= 3/4 -(1/4)(2n + 3) .(1/3)^n
S = 1.(1/3)^1 + 2.(1/3)^2+....+n.(1/3)^n (1)
把等式（1） * 1/3 得到等式（2）
(1/3)S = 1.(1/3)^2 + 2.(1/3)^3+....+n.(1/3)^(n+1) (2)
(1)-(2)
(2/3) S
=[ 1.(1/3)^1 + 2.(1/3)^2+....+n.(1/3)^n ] -[ 1.(1/3)^2 + 2.(1/3)^3+....+n.(1/3)^(n+1)]
= 1/3 +[2.(1/3)^2 -1.(1/3)^2]+[3.(1/3)^3 -2.(1/3)^3]+...+[ n.(1/3)^n - (n-1)(1/3)^n]
-n.(1/3)^(n+1)
=( 1/3 + 1/3^2+...+1/3^n) -n.(1/3)^(n+1)
=(1/2) ( 1- (1/3)^n ) - n.(1/3)^(n+1)
S = 3/4 - (1/4)(2n + 3) .(1/3)^n追问谢谢。

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