等差数列an的前n项和胃Sn,bn=1/Sn,且a3b3=1/2,S3+S5=21.求证b1+b2+b3...+bn

题目

等差数列an的前n项和胃Sn,bn=1/Sn,且a3b3=1/2,S3+S5=21.求证b1+b2+b3...+bn

答案

a3*b3=a3/s3=1/2,即2a3=a1+a2+a3,∴a1=d ...①s3+s5=21,即3a1+(3*2d/2)+5a1+(5*4d/2)=21...②由①②得a1=d=1,∴an=n,sn=n(1+n)/2∴bn=2/n(1+n)=2[(1/n)-(1/(n+1) )]∴b1+b2+...+bn=2[1-(1/(n+1)]=2-[2/(n+1)]...