已知数列an满足an+1=7an-2/2an+3,a1=2,求数列an的通项公式

Question

已知数列an满足an+1=7an-2/2an+3,a1=2,求数列an的通项公式

Answer 1

a(n+1)=(7an -2)/(2an +3)
a(n+1) -1=(7an-2-2an-3)]/(2an +3)=(5an-5)/(2an+3)=5(an-1)/(2an+3)
1/[a(n+1)-1]=(2an+3)/[5(an-1)]=(2an-2+5)/[5(an-1)]=1/(an -1) +2/5
1/[a(n+1)-1]-1/(an -1)=2/5,为定值.
1/(a1-1)=1/(2-1)=1,数列{1/(an -1)}是以1为首项,2/5为公差的等差数列.
1/(an -1)=1+(2/5)(n-1)=(2n+3)/5
an=(2n+8)/(2n+3)=1+ 5/(2n+3)
数列{an}的通项公式为an=1+ 5/(2n+3).