Question

(1/2)+[3/(2^2)]+[5/(2^3)]+...+[(2n-1)/(2^n)] 的前 n 项和 Sn 为 __

Answer 1

sn=1/2+3/2^2+5/2^3+...+(2n-1)/2^n (1)

2*sn=1+3/2+5/2^2+...+(2n-1)/2^(n-1) (2)

由 (2)-(1)得

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

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

=3-(2n+3)/2^n.