设数列{an}的前n项和为sn,若对于任意的正整数n都有sn=2an-3n.(1)设bn=an+3,证明:数列{bn}是等比数列百度里的答案an=Sn-S(n-1)=2an-3n-[2a(n-1)-3(n-1)]=2an-2a(n-1)-3即:an=2a(n-1)+3 为什么2an-2a(n-1)-3等于an
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![设数列{an}的前n项和为sn,若对于任意的正整数n都有sn=2an-3n.(1)设bn=an+3,证明:数列{bn}是等比数列百度里的答案an=Sn-S(n-1)=2an-3n-[2a(n-1)-3(n-1)]=2an-2a(n-1)-3即:an=2a(n-1)+3 为什么2an-2a(n-1)-3等于an](/uploads/image/z/1740906-18-6.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E8%8B%A5%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8F%E7%9A%84%E6%AD%A3%E6%95%B4%E6%95%B0n%E9%83%BD%E6%9C%89sn%3D2an-3n.%281%29%E8%AE%BEbn%3Dan%2B3%2C%E8%AF%81%E6%98%8E%3A%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%E7%99%BE%E5%BA%A6%E9%87%8C%E7%9A%84%E7%AD%94%E6%A1%88an%3DSn-S%28n-1%29%EF%BC%9D2an-3n-%5B2a%28n-1%29-3%28n-1%29%5D%EF%BC%9D2an-2a%28n-1%29-3%E5%8D%B3%EF%BC%9Aan%EF%BC%9D2a%28n-1%29%2B3+%E4%B8%BA%E4%BB%80%E4%B9%882an-2a%28n-1%29-3%E7%AD%89%E4%BA%8Ean)
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