数列求和an等于n的平方除以2的n次方，，，则sn怎么求

Question

如题所述

Answer 1

这里用2次错项相减法

原式①*2=1+2*2/2+3*3/2^2+…+n*n/2^（n-1）②
②-①=1+3/2+5/4+…+(2n-1)/2^(n-1)-n*n/2^n③
这里再对③用同样步骤，过程不详细写了
答案6-(n*n+4n+6)/2^n本回答被提问者采纳

Answer 2

consider
n^2-(n-1)^2 = 2n-1

let
S1 = 1.(1/2)^1+2(1/2)^2+...+n.(1/2)^n (1)
(1/2)S1 = 1.(1/2)^2+2(1/2)^3+...+n.(1/2)^(n+1) (2)
(1)-(2)
(1/2)S1 = (1/2+1/2^2+...+1/2^n) -n.(1/2)^(n+1)
= [1-(1/2)^n] -n.(1/2)^(n+1)
S1 = 2[1-(1/2)^n] -n.(1/2)^n

let
cn = (2n-1)(1/2)^n
= 2n(1/2)^n - (1/2)^n
Tn =c1+c2+....+cn
= 2S1 -[1-(1/2)^n]
= 3[1-(1/2)^n] -2n.(1/2)^n
= 3-(2n+3).(1/2)^n

an= n^2/2^n
let

S = 1^2.(1/2)^1 + 2^2.(1/2)^2+....+n^2.(1/2)^n (3)
(1/2)S = 1^2.(1/2)^2 + 2^2.(1/2)^3+....+n^2.(1/2)^(n+1) (4)
(3)-(4)
(1/2)S ={ [2(1)-1](1/2)^1 + [2(2)-1](1/2)^2 +...+ (2n-1)(1/2)^n } - n^2.(1/2)^(n+1)
= Tn -n^2.(1/2)^(n+1)
=3-(2n+3).(1/2)^n - n^2.(1/2)^(n+1)
S = 6-2(2n+3).(1/2)^n - n^2.(1/2)^n
=6-(n^2+4n+6).(1/2)^n

a1+a2+...+an =S
= 6-(n^2+4n+6).(1/2)^n

Answer 3

先求2sn，再2sn-sn