令bn＝an－2
an-2=1/2an-1-1=1/2(an-1-2)
∴(an-2)/(an-1-2)=1/2
即bn/bn-1=1/2
又有a1=1 b1=-1
∴｛bn｝为首项b1=-1,公比是1/2的等比数列
bn｝通项公式为bn=-(1/2)^(n-1)
带入bn＝an－2得 an=2-(1/2)^(n-1) (n≥2)
a=1时不成立 此时a1=1
数列{an}满足a1=1,an=1/2an-1+1(n≥2)
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