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)
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
1个回答
相关问题
-
数列an的前n项和Sn,a1=1,a(n+1)=2Sn.(1).求数列an的通项公式 (2)求数列nan的前n项和Tn.
-
等差数列{an}前n项和Sn,{bn}前n项和Tn,有Sn/Tn=(4n+1)/(3n-
-
设数列an前n项和为Sn,且Sn=2^n-1 1.求an通项公式 2.设bn=n*an求bn前n项和Tn
-
等差数列{an}和{bn}前n项和为sn,Tn且sn/tn=2n+1/n+2求a5/b5
-
等差数列{an}、{bn}的前n项和分别为Sn、Tn,若Sn/Tn=2n/3n+1,求an/bn
-
设等差数列{an}的前n项和为Sn,等差数列{bn}的前n项和为Tn,若Tn\Sn=4n+27\7n+1,求bn\an
-
已知数列{an}前n项和为Sn,a1=1,an+1=2Sn,求{nan}的前n项和Tn.
-
等差数列前几项和SnTn,Sn/Tn=2n/3n+1求an/bn
-
等差数列{an}的前n项和Sn=4n2-25n.求数列{|an|}的前n项的和Tn.
-
等差数列{an}的前n项和Sn=4n2-25n.求数列{|an|}的前n项的和Tn.