求数列|nx(1/2n)|的前n项和
1个回答

感觉应该是nx(1/2^n)如果是的话,就用错位相减法

S=1.(1/2)^1+2(1/2)^2+3(1/2)^3..……+(n-1)(1/2)^(n-1)+n.(1/2)^n (1)

(1/2)S= 1.(1/2)^2+2.(1/2)^3+...……………… +(n-1)(1/2)^n+n(1/2)^(n+1) (2)

(1)-(2)得

(1/2)S = (1/2+1/2^2+...+1/2^n) -n.(1/2)^(n+1)

= (1- (1/2)^n)-n.(1/2)^(n+1)

S =2(1- (1/2)^n)-n.(1/2)^n

= 2 - (n+2).(1/2)^n