Question

在等差数列{an}中,Sn是其前n项的和,且a1=2,S2009/2009-S2007/2007=2，则数列｛1/Sn｝的前n项的和是

Answer 1

S2009/2009 -S2007/2007
=[2009a1+2009×2008d/2]/2009 -[2007a1+2007×2006d/2]/2007
=a1+1004d -a1-1003d
=d=2
Sn=na1+n(n-1)d/2=2n+n(n-1)=n(n+1)
1/Sn=1/[n(n+1)]=1/n -1/(n+1)
前n项和
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)

Answer 2

Sn=na1+n(n-1)d/2
所以Sn/n=a1+(n-1)d/2=2+(n-1)d/2
所以S2009/2009=2+1004d
S2007/2007=2+1003d
所以S2009/2009-S2007/2007=(2+1004d)-(2+1003d)=d=2
所以Sn=na1+n(n-1)d/2=2n+n(n-1)=n(n+1)
所以1/Sn=1/[n(n+1)]=1/n-1/(n+1)
所以数列{1/Sn}的前n项的和Tn=1-1/2+1/2-1/3+…+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)

Answer 3

n/(n+1)