等差数列{an}中,a1=3,公差d=2,Sn为前n项和,求1/S1+1/S2+…+1/Sn

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等差数列{an}中,a1=3,公差d=2,Sn为前n项和,求1/S1+1/S2+…+1/Sn等差数列{an}中,a1=3,公差d=2,Sn为前n项和,求1/S1+1/S2+…+1/Sn等差数列{an}

等差数列{an}中,a1=3,公差d=2,Sn为前n项和,求1/S1+1/S2+…+1/Sn
等差数列{an}中,a1=3,公差d=2,Sn为前n项和,求1/S1+1/S2+…+1/Sn

等差数列{an}中,a1=3,公差d=2,Sn为前n项和,求1/S1+1/S2+…+1/Sn
又题知:a n=2n+1 ∴S n=n2+2n=n(n+2) ∴1/Sn=1/2〔1/n-/1(n+2)〕
∴1/S1+1/S2+...+1/Sn=1/2(1-1/3+1/2-1/4+1/3-1/5+...+1/n-1 -1/n+1+1/n-1/n+2)
=1/2 (3/2-1/n+1-1/n+2)=3/4-1/2(n+1)-1/2(n+2) 你自己化解

an=2n+1,sn=n(n+2), 1/sn=1/2(1/n-1/(n+2))
所以 1/S1+1/S2+…+1/Sn =1/2(1-1/3+1/2-1/4+1/3-1/5.....-1/(n+1)-1/(n+2))=3/4-(2n+3)/[2(n+1)(n+2)]

3/4-(2n+3)/[2(n+1)(n+2)]