赞 扭扭的妞妞 幼苗
共回答了23个问题采纳率：87% 举报
2/2,4/2^2,6/2^3,...,2n/2^n
let
S = 1.(1/2)^0+2.(1/2)^1+...+n.(1/2)^(n-1) (1)
(1/2)S = 1.(1/2)^1+2.(1/2)^2+...+n.(1/2)^n (2)
(1)-(2)
(1/2)S = [1+ 1/2+...+1/2^(n-1)] -n.(1/2)^n
= 2(1- (1/2)^n) -n.(1/2)^n
= 2 - (n+2)(1/2)^n
S = 4-2(n+2)(1/2)^n
an = 2n/2^n
= n(1/2)^(n-1)
Sn =b1+b2+...+bn
=S
=4-2(n+2)(1/2)^n
let
S = 1.(1/2)^0+2.(1/2)^1+...+n.(1/2)^(n-1) (1)
(1/2)S = 1.(1/2)^1+2.(1/2)^2+...+n.(1/2)^n (2)
(1)-(2)
(1/2)S = [1+ 1/2+...+1/2^(n-1)] -n.(1/2)^n
= 2(1- (1/2)^n) -n.(1/2)^n
= 2 - (n+2)(1/2)^n
S = 4-2(n+2)(1/2)^n
an = 2n/2^n
= n(1/2)^(n-1)
Sn =b1+b2+...+bn
=S
=4-2(n+2)(1/2)^n
1年前
9
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