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≥2时,a[n]=s[n]-s[n-1]
=1/(2n-1)-1/(2n-3)
=-2/[(2n-1)(2n-3)]
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≥2时,a[n]=s[n]-s[n-1]
=1/(2n-1)-1/(2n-3)
=-2/[(2n-1)(2n-3)]