等差数列{an}a1=2,d=2,求前n项和Sn以及求通向公式{1/Sn}的前n项和Tn

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Sn=nA1+(1/2)n(n-1)d=2n+n(n-1)=n(n+1)1/Sn=1/[n(n+1)]=[(n+1)-n]/[n(n+1)]=1/n-1/(n+1)Tn=1/S1+1/S2+……+1/Sn=(1/1-1/2)+(1/2-1/3)+……+(1/n-1/(n+1))=1-1/(n+1)=n/(n+1)