两边同时除以sn得1=1/2(sn-1)+1/sn
设1/sn-a=-1/2(1/(sn-1)-a)
解得a=2/3,又a1=2,所以1/s1-2/3=-1/6
所以1/sn-2/3=（-1/6）（-1/2）^(n-1)
得1/sn=1/3*（-1/2）^n+2/3
1/s(n+1)-1/sn=（-1/2）^(n+1)不等于常数
所以数列｛1/sn｝不是等差数列
已知数列{an}的前几项和为sn,且满足sn=sn/2s(n-1)+1,(a≥2),a1=2,求{1/sn}是等差数列
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