已知f(x)=3x/(x+3),数列an满足an=f(an-1)(n>1,n属于N*,a1不等于0,求数列an的通项公式.

Question

已知f(x)=3x/(x+3),数列an满足an=f(an-1)(n>1,n属于N*,a1不等于0,求数列an的通项公式.

Answer 1

n>1时,an=f[a(n-1)]=3a(n-1)/[a(n-1)+3]
1/an=[a(n-1)+3]/[3a(n-1)]=1/a(n-1) +1/3
1/an -1/a(n-1)=1/3,为定值.
数列{1/an}是以1/a1为首项,1/3为公差的等差数列.
1/an =1/a1 +(1/3)(n-1)=1/a1 +(n-1)/3=[3+(n-1)a1]/(3a1)
an=(3a1)/[3+(n-1)a1]