已知数列a1=1,an=a(n-1)/3a(n-1)+1(n>=2)设bn=ana(n+1),求数列{an}的通项公式求数列{bn}的前n项和sn
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An=[A(n-1)]/[3A(n-1)+1]
==> 1/An =3 +1/A(n-1)
==> {1/an}为等差数列, 首项 =1/A1 =1, 公差 =3
1/An =1/A1 +3(n-1) =3n-2
==> An =1/(3n-2)
Bn =An*A(n+1) =1/(3n-2)(3n+1) =[1/(3n-2) -1/(3n+1)]/3
==> Sn =[1 -1/(3n+1)]/3 = n/(3n+1)
(1)
an=a(n-1)/[3a(n-1)+1],
则
1/an = [3a(n-1)+1]/a(n-1) = 3 + 1/a(n-1)
说明 数列{1/an}是等差数列,公差为 3
1/an = 1/a1 + 3(n-1) = 3n -2
所以 an = 1/(3n-2)
(2) bn=ana(n+1)= 1/(3n-2)* 1/(3n+1)= [1/(3n-2)- 1/(3n+1)]/ 3
数列{bn}的前n项和sn = [1/(3-2)- 1/(3n+1)]/ 3 = n/(3n+1)
由an=a(n-1)/3a(n-1)+1
得1/an=1/a(n-1)+3
令cn=1/an
cn=c(n-1)+3
cn=1+3(n-1)=3n-2
an=1/(3n-2)
bn=1/(3n-2)*1/(3n+1)=[1/(3n-2)-1/(3n+1)]/3=[an-a(n+1)]/3
Sn=[a1-a(n+1)]/3=n/(3n+1)
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