n≥2时,a[n]=s[n]-s[n-1],将它代入an= 2Sn^2/2Sn-1 ,并化简,得1/s[n]=1/s[n-1]+2 （n≥2）上式表明{1/s[n]}是以1/s[1]=1/a[1]=1 为首项,2为公差的等差数列所以1/s[n]=2n-1,s[n]=1/(2n-1) （n≥1）故n=1时,a[1]=1; n...