Question

已知等差数列an的前n项和为Sn,bn=1/Sn,且a3b3=1/2,S3+S5=21

Answer 1

1.

a3b3=a3/S3=1/2,S3=a1+a2+a3

故：a1+a2+a3=2a3 得：a1=d

又：S3+S5=21 解出：a1=d=1,an=a1+(n-1)d=n

Sn=(a1+an)n/2=(an+1)n/2=(n+1)n/2

bn=2/n(n+1)

2.bn=2/n(n+1)=2*[1/n(n+1)]=2*[(1/n-1/(n+1)]

b1+b2+.bn=2*[1-1/2+1/2-1/3+...+1/n-1/(n+1)]=2*[1-1/(n+1)]