t=1/2+3/2^2+5/2^3+...+(2n-3)/2^(n-1)+(2n-1)/2^n
2t=1+3/2+5/2^2+...+(2n-3)/2^(n-2)+(2n-1)/2^(n-1)
相减有
t=1+[2/2+2/2^2+2/2^3+...2/2^(n-1)]-(2n-1)/2^n
=1+[(1/2)^(n-1)-1]/(1/2-1)]-(2n-1)/2^n
=1+2-2^(2-n)-(2n-1)/2^n
=3-(2n+3)/2^n