设{an}为等比数列,a1=1,a2=3,求最小的自然数n,使an>=2007; (2),求和:T2n=1/a1-2/a2+3/a3-…-2n/a2n
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![设{an}为等比数列,a1=1,a2=3,求最小的自然数n,使an>=2007; (2),求和:T2n=1/a1-2/a2+3/a3-…-2n/a2n](/uploads/image/z/15232500-36-0.jpg?t=%E8%AE%BE%7Ban%7D%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2Ca1%3D1%2Ca2%3D3%2C%E6%B1%82%E6%9C%80%E5%B0%8F%E7%9A%84%E8%87%AA%E7%84%B6%E6%95%B0n%2C%E4%BD%BFan%3E%3D2007%3B+%282%29%2C%E6%B1%82%E5%92%8C%EF%BC%9AT2n%3D1%2Fa1-2%2Fa2%2B3%2Fa3-%E2%80%A6-2n%2Fa2n)
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