Question

数列 (24 10:6:29) 已知等差数列｛an｝的前n项和Sn,bn=1/Sn,且a3b3=1/2,S3+

Answer 1

1)设an=a1+(n-1)d,则Sn=na1+n(n-1)d/2

由a3b3=1/2,bn=1/Sn,得S3=2a3=2a1+4d=3a1+3d,a1=d

所以Sn=n(n+1)d/2

由S3+S5=21,得6d+15d=21,d=1

所以Sn=n(n+1)/2

所以：bn=2/[n(n+1)]

2)b1+b2+b3+……+bn

=2{1/[1*2]+1/[2*3]+……+1/[n*(n+1)]}

=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]

=2[1-1/(n+1)]