数列an有a1=1,n>=2时,3tSn-(2t+3)S(n-1)=3t(常数a>0),问:求an的通项公式问题二若a(n+1)=an·f(t),bn=f(1/(n-1),求bn问题三求和b1b2-b2b3+b3b4-b4b5+…-b2n·b(2n+1)
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![数列an有a1=1,n>=2时,3tSn-(2t+3)S(n-1)=3t(常数a>0),问:求an的通项公式问题二若a(n+1)=an·f(t),bn=f(1/(n-1),求bn问题三求和b1b2-b2b3+b3b4-b4b5+…-b2n·b(2n+1)](/uploads/image/z/7658997-69-7.jpg?t=%E6%95%B0%E5%88%97an%E6%9C%89a1%3D1%2Cn%3E%3D2%E6%97%B6%2C3tSn-%282t%2B3%29S%28n-1%29%3D3t%28%E5%B8%B8%E6%95%B0a%3E0%29%2C%E9%97%AE%EF%BC%9A%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E9%97%AE%E9%A2%98%E4%BA%8C%E8%8B%A5a%28n%2B1%29%3Dan%C2%B7f%28t%29%EF%BC%8Cbn%3Df%281%2F%28n-1%29%2C%E6%B1%82bn%E9%97%AE%E9%A2%98%E4%B8%89%E6%B1%82%E5%92%8Cb1b2-b2b3%2Bb3b4-b4b5%2B%E2%80%A6-b2n%C2%B7b%282n%2B1%29)
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