如果你题目是这样的,s(n+1)sn=a(n+1),那么s(n+1)sn=a(n+1)=s(n+1)-sn两边除以s(n+1)sn得1=1/sn-1/s(n+1),即1/s(n+1)=1/sn-1s1=a1=2所以1/sn是首项为1/s1=1/2,公差为-1的等差数列1/sn=(3-2n)/2所以sn=2/(3-2n)...