∵点P(an,an+1)在直线x-y+2=0上∴an-a(n+1)+2=0a(n+1)=an+2∴an=a1+2(n-1)=2n∴Sn=2×n(n+1)/2=n²+n=n(n+1)∴1/Sn=1/n(n+1)=1/n-1/(n+1)∴1/S1+1/S2+1/S3..+1/Sn=1-1/2+1/2-1/3+1/3-1/4..+1/n-1/(n+1)=1-1/(n+1...