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an+Sn=2n 令n=1 a1+S1=2=>a1=1 又a(n-1)+S(n-1)=2(n-1) 與上式作差 an-a(n-1)+an=2 2an-a(n-1)=2 an-2=(1/2)[a(n-1)-2] 得證 an-2首項a1-2=-1 公比1/2 an=-1/2^(n-1)
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