Question

已知数列{an}满足a1=5,a(n+1)=(8an-12)/(3an-4),bn=1/(an-

)

Answer 1

a(n+1)=(8an-12)/(3an-4)
a(n+1)-2=(8an-12-6an+8)/(3an-4)=(2an-4)/(3an -4)
1/[a(n+1)-2]=(3an -4)/(2an -4)
=(1/2)[(3an-4)/(an-2)]
=(1/2)[(3an-6+2)/(an -2)]
=(1/2)[3 +2/(an -2)]
=1/(an -2) +3/2
1/[a(n+1)-2]-1/(an -2)=3/2，为定值
1/(a1-2)=1/(5-2)=1/3，数列{1/(an -2)}是以1/3为首项，3/2为公差的等差数列
bn=1/(an -2)，数列{bn}是以1/3为首项，3/2为公差的等差数列
1/(an -2)=1/3+(3/2)(n-1)=(9n-7)/6
an=6/(9n-7) +2=2(9n-4)/(9n-7)
n=1时，a1=2(9×1-4)/(9×1-7)=5，同样满足通项公式
数列{an}的通项公式为an=2(9n-4)/(9n-7)