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=n1/(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)