设原式=Sn2*Sn=1+3/2+5/4+7/8+...+(2n-1)/(2^(n-1))上式减去原式（错位相减）得出Sn=1+2*1/2+2*1/4+2*1/8+...+2*1/(2^(n-1))-(2n-1)/(2^n)=1-(2n-1)/(2^n)+[1+1/2+1/4+...+1/(2^(n-2))]=1-(2n-1)/(2^n)+2-1/(2^(...