已知等差数列｛an｝的前n项之和为Sn,且S3+S5=21.设bn=1/Sn,a3b3=1/2,求｛bn｝的前n项之和为Bn
问题描述：
已知等差数列｛an｝的前n项之和为Sn,且S3+S5=21.设bn=1/Sn,a3b3=1/2,求｛bn｝的前n项之和为Bn
答
b3=1/S3=1/(a1+a2+a3)=1/(3a1+3d)
a3b3=1/2=(a1+2d)/3(a1+d)
2(a1+2d)=3(a1+d)
a1=d
3a1+3d+5a1+10d=21
8a1+13d=21
8d+13d=21
d=1
a1=d=1
an=1+(n-1)*1=n
a2=2
a3=3
.
sn=(1+n)*n/2
bn=1/sn=2/(1+n)*n
b1=1/1*2
b2=2/2*3
b3=2/3*4
.
bn=2/(1+n)*n
Sbn=2/1*2+2/2*3+2/3*4+.+2/n(n+1)
=2[1/1*2+1/2*3+1/3*4+.+1/n(n+1)]
=2[1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)