Question

设等差数列{an}的前n项和为Sn,bn=1/Sn,且a3b3=1/2,S5+S3=21,求bn.

Answer 1

解：
设等差数列{an}公差为d。
a3b3=a3/S3=a3/(3a1+3d)=a3/[3(a1+d)]=a3/(3a2)=1/2
a3=(3/2)a2=a2+d
d=a2/2
S5+S3
=5a1+10d+3a1+3d
=5(a2-d)+10d+3(a2-d)+3d
=8a2+5d=8a2+(5/2)a2
=(21/2)a2
=21
a2=2
d=a2/2=2/2=1
a1=a2-d=2-1=1
an=1+n-1=n
Sn=1+2+...+n=n(n+1)/2
bn=2/[n(n+1)]=2[1/n -1/(n+1)]

Answer 2

等差数列{an}的前n项和为Sn,bn=1/Sn,且a3b3=1/2
a3b3=a3/S3=a3/(a1+a2a+a3)=1/2
∴a1+a2=a3,公差d=a1,an=n·a1
S5+S3=15a1+6a1=21a1=21
∴a1=1,an=n,Sn=n(n+1)/2,bn=2/n(n+1)
第二题a1=1,a2=2a1/(a1+2)=2/3,a3=2a2/(a2+2)=1/2
1/an=[a(n-1)+2]/2a(n-1)=1/2+1/a(n-1)
=1/2+1/2+1/a(n-2)
∴｛1/an｝成等差数列
1/an=(1+n)/2,an=2/(1+n)

Answer 3

(1)等差数列{an}的前n项和为Sn,bn=1/Sn,且a3b3=1/2
a3b3=a3/S3=a3/(a1+a2a+a3)=1/2
∴a1+a2=a3,公差d=a1,an=n·a1
S5+S3=15a1+6a1=21a1=21
∴a1=1,an=n,Sn=n(n+1)/2,bn=2/n(n+1)
(2)b1=1,b2=1/3
Tn估计是bn前n项之和吧，
Tn=2/[(1+n)n]
=2*[1/1-1/2+1/2-1/3+1/3-1/4....1/n-1/(n-1)]
=2*(1-1/(n+1))
=2-2/(n+1)
T2=4/3
第二题
(1)an=2a(n-1)/[a(n-1)+2]两边分别去倒数，
1/an=1/2+1/a(n-1)
1/an-1/a(n-1)=1/2
所以｛1/an｝成等差数列
(2)设Cn=1/an,所以Cn为成等差数列，C1=1/a1=1，公差d=1/an-1/a(n-1)=1/2
所Cn=1+(n-1)/2
所以1/an=1+(n-1)/2
an=2/（n+1）