数列{an}满足a1=1,an+1=（an+3）除以（2an-4）求an的通项公式?

Question

数列{an}满足a1=1,an+1=（an+3）除以（2an-4）求an的通项公式?

Answer 1

使用不动点求解该类数列题目!
考虑方程x=(x+3)/(2x-4)的x1=-1/2,x2=3
∴a(n+1)-3=(an+3)/(2an-4)-3=(15-5an)/(2an-4)
a(n+1)+1/2=(an+3)/(2an-4)+1/2=(2an+1)/(2an-4)
∴两式相除,得：[a(n+1)-3]/[a(n+1)+1/2]=(15-5an)/(2an+1)=-5/2*(an-3)/(an+1/2)
∴数列{(an-3)/(an+1/2)}为等比数列,
∴(an-3)/(an+1/2)=(-4/3)*(-5/2)^(n-1)
∴解得：an=7/[2(1-(-4/3)*(-5/2)^(n-1))]-1/2