数列﹛an﹜的前n项和Sn满足﹙a-1﹚Sn=a﹙an-1﹚数列﹛bn﹜满足bn=an•lg an请讲下那个an
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![数列﹛an﹜的前n项和Sn满足﹙a-1﹚Sn=a﹙an-1﹚数列﹛bn﹜满足bn=an•lg an请讲下那个an](/uploads/image/z/7163565-69-5.jpg?t=%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3%EF%B9%99a%EF%BC%8D1%EF%B9%9ASn%EF%BC%9Da%EF%B9%99an%EF%BC%8D1%EF%B9%9A%E6%95%B0%E5%88%97%EF%B9%9Bbn%EF%B9%9C%E6%BB%A1%E8%B6%B3bn%EF%BC%9Dan%26%238226%3Blg+an%E8%AF%B7%E8%AE%B2%E4%B8%8B%E9%82%A3%E4%B8%AAan)
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数列﹛an﹜的前n项和Sn满足﹙a-1﹚Sn=a﹙an-1﹚数列﹛bn﹜满足bn=an•lg an请讲下那个an
数列﹛an﹜的前n项和Sn满足﹙a-1﹚Sn=a﹙an-1﹚数列﹛bn﹜满足bn=an•lg an
请讲下那个an
数列﹛an﹜的前n项和Sn满足﹙a-1﹚Sn=a﹙an-1﹚数列﹛bn﹜满足bn=an•lg an请讲下那个an
已知Sn满足﹙a-1﹚Sn=a﹙an-1﹚……(1)
则﹙a-1﹚Sn-1=a﹙an-1-1﹚……(2)
(1)—(2)得﹙a-1﹚(Sn—Sn-1)=a﹙an-an-1﹚
即﹙a-1)an=a﹙an-an-1﹚→an/an-1=a
由(1)式,令n=1得a1=a
∴an为等比数列,且an=a×a^(n-1)=a^n