Question

已知数列an中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于N*)在直线x-y+1=0上,则1/S1+1/S2+1/S3+……+1/S99=

Answer 1

∵(an,an+1)(n属于N*)在直线x-y+1=0上
∴an+1-an=1
∴a1=1,d=1的等差数列
∴an=1+(n-1)x1=n
1/S1+1/S2+1/S3+……+1/S99
=1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+99)
=2/1x2+2/2x3+2/3x4+...+2/99x100
=2(1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100)
=2(1-1/100)=99/50

Answer 2

易得an=n,所以Sn=[n(n+1)]/2,所以1/Sn=2/[n(n+1)]=2[1/n-1/(n+1)].所以1/S1+1/S2+1/S3+…+1/S99=2*(1-1/2+1/2-1/3+1/3-1/4+…+1/99-1/100)=99/50