a3=a1+2d=6 S3=a1+a2+a3=3a1+3d=12 解得a1=2,d=2,故an=2n 所以Sn=n(n+1) 所以1/S1+1/S2+……+1/Sn =1/(1*2)+1/(2*3)+1/(3*4)+……+1/n(n+1) =(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+[1/n-1/(n+1)] =1-1/(n+1)
已知数列an是等差数列,其前n项和为Sn,a3=6,
1个回答
相关问题
-
已知数列{an}中,其前n项和为Sn,且n,an,Sn成等差数列(n∈N*).
-
已知数列{an}中,其前n项和为Sn,且n,an,Sn成等差数列(n∈N*).
-
已知数列{an}是等差数列,其前n项和为Sn,a3=7,S4=24
-
已知数列{an}是等差数列,其前n项和为Sn,a3=7,S4=24.
-
已知数列{An}是等差数列前n项和Sn,A3=6,S3=12.
-
已知数列{an}为等差数列,Sn为其前n项和,a1+a5=6,S9=63.
-
已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.
-
已知等差数列{an}的前n项和为sn,求证数列{sn/n}也成等差数列
-
已知数列﹛an﹜是等差数列且a1=12 a6=27求数列﹛an+2^n﹜的前n项和Sn
-
已知数列an的前n项和为sn=5/6n(n+3),1:求证an为等差数列 2:设bn=a3n+a