急求!高一数学题:已知数列{an},a1 = 1 , Sn是前n项和,Sn+1= Sn/( 3+4Sn) n >= 1 , 求an通项公式
人气：221 ℃ 时间：2019-08-21 13:02:37
解答
1/S(n+1) = 3/Sn + 4令1/Sn = bn则有b(n+1) = 3bn +4b(n+1) + 2 = 3(bn + 2)等比数列,则bn+2 = (b1+2)*3^(n-1)b1=1/S1=1/a1=1所以bn=3^n - 2Sn=1/(3^n-2)Sn-1=1/(3^(n-1)-2)an=Sn-Sn-1=.
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