数列an的前n项和为sn,且满足a1=1,2Sn=(n+1)an (1)求{an}的通项公式(2)求和Tn=1/2a1+
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数列an的前n项和为sn,且满足a1=1,2Sn=(n+1)an (1)求{an}的通项公式(2)求和Tn=1/2a1+1/3a2+……+1/(n+1)an
![数列an的前n项和为sn,且满足a1=1,2Sn=(n+1)an (1)求{an}的通项公式(2)求和Tn=1/2a1+](/uploads/image/z/4088545-25-5.jpg?t=%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E4%B8%94%E6%BB%A1%E8%B6%B3a1%3D1%2C2Sn%3D%28n%2B1%29an+%281%29%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%282%29%E6%B1%82%E5%92%8CTn%3D1%2F2a1%2B)
∵2sn=(n+1)an
∴2s(n-1)=na(n-1)
两式相减:
∴an=n[an-a(n-1)]
即an/a(n-1)=n/(n-1)
∴an=n
1/(n+1)an=1/n(n+1)=1/n-1/(n+1)
∴Tn=(1-1/2)+(1/2-1/3)+……(1/n-1/n+1)
=1-1/n+1
=n/(n+1)
∴2s(n-1)=na(n-1)
两式相减:
∴an=n[an-a(n-1)]
即an/a(n-1)=n/(n-1)
∴an=n
1/(n+1)an=1/n(n+1)=1/n-1/(n+1)
∴Tn=(1-1/2)+(1/2-1/3)+……(1/n-1/n+1)
=1-1/n+1
=n/(n+1)
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