设二次方程anx^2-a(n+1)x+1=0有两个根x1,x2,且满足6x1-2x1x2+6x2=3.已知a1=7/6.(1)用an表示a(n+1)(2)求{an}的通项公式(3)求{an}的前n项之和sn
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![设二次方程anx^2-a(n+1)x+1=0有两个根x1,x2,且满足6x1-2x1x2+6x2=3.已知a1=7/6.(1)用an表示a(n+1)(2)求{an}的通项公式(3)求{an}的前n项之和sn](/uploads/image/z/980874-18-4.jpg?t=%E8%AE%BE%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Banx%5E2-a%28n%2B1%29x%2B1%3D0%E6%9C%89%E4%B8%A4%E4%B8%AA%E6%A0%B9x1%2Cx2%2C%E4%B8%94%E6%BB%A1%E8%B6%B36x1-2x1x2%2B6x2%3D3.%E5%B7%B2%E7%9F%A5a1%3D7%2F6.%281%29%E7%94%A8an%E8%A1%A8%E7%A4%BAa%28n%2B1%29%282%29%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%283%29%E6%B1%82%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E4%B9%8B%E5%92%8Csn)
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