等差数列{an}=2n+1,前n项和为Sn,求1/S1+1/S2+.+1/Sn

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等差数列{an}=2n+1,前n项和为Sn,求1/S1+1/S2+.+1/Sn等差数列{an}=2n+1,前n项和为Sn,求1/S1+1/S2+.+1/Sn等差数列{an}=2n+1,前n项和为Sn,

等差数列{an}=2n+1,前n项和为Sn,求1/S1+1/S2+.+1/Sn
等差数列{an}=2n+1,前n项和为Sn,求1/S1+1/S2+.+1/Sn

等差数列{an}=2n+1,前n项和为Sn,求1/S1+1/S2+.+1/Sn
an=2n+1
Sn=n(a1+an)/2=n(3+2n+1)/2=n(n+2)
所以1/Sn=1/n(n+2)=[1/n-1/(n+2)]/2
那么1/S1+1/S2+.+1/Sn
=(1-1/3)/2+(1/2-1/4)/2+(1/3-1/5)/2+...+[1/n-1/(n+2)]/2
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-(2n+3)/2(n+1)(n+2)
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由an求出sn=n(n+2)
1/sn=1/n-1/(n+2)
1/S1+1/S2+.....+1/Sn=1-1/(n+2)

an=2n+1
Sn=(a1+an)n/2=n^2+2n
1/S1+1/S2+.....+1/Sn
=1/(1x3)+1/(2x4)+……+1/[n(n+2)]
=1/2[(1-1/3)+1/2-1/4)+……+1/n-1/(n+2)]
=1/2[1+1/2-1/n-1(n+2)]
3/2-(n+1)/n(n+2)

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