S(n+1)²-Sn²-S(n+1)-3Sn-a(n+1)=0可化为:S(n+1)²-Sn²-S(n+1)-3Sn-[S(n+1)-Sn]=0S(n+1)²-Sn²=2[S(n+1)-Sn][S(n+1)+Sn][S(n+1)-Sn]=2[S(n+1)-Sn](1)如果[S(n+1)-Sn]=0==》a(n+1)=0(n≥...
{an中},a1=1,S(n+1)²-Sn²-S(n+1)-3Sn-a(n+1)=0,求an
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