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

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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=1,2S

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+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+1/3a2+……+1/(n+1)an
∵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)

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