设a1=5,当n≥2时an+an-1=7/(an-an-1)+6,求数列(an)的通项公式
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![设a1=5,当n≥2时an+an-1=7/(an-an-1)+6,求数列(an)的通项公式](/uploads/image/z/2695412-20-2.jpg?t=%E8%AE%BEa1%3D5%2C%E5%BD%93n%E2%89%A52%E6%97%B6an%2Ban-1%3D7%2F%28an-an-1%29%2B6%2C%E6%B1%82%E6%95%B0%E5%88%97%28an%29%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
设a1=5,当n≥2时an+an-1=7/(an-an-1)+6,求数列(an)的通项公式
设a1=5,当n≥2时an+an-1=7/(an-an-1)+6,求数列(an)的通项公式
设a1=5,当n≥2时an+an-1=7/(an-an-1)+6,求数列(an)的通项公式
最后结果我有点不太确定(因为时间长没做这种题了),但具体方法就是上面所写的.