数列难题,若F（X)=(3X+2)/(X+2) 满足A1=0.5 A(n+1)=F(An) B(n)=1/(An +1)

a(n+1)=(3*an+2)/(an+2)上式两边同加1的a(n+1)+1=(3*an+2)/(an+2)+1a(n+1)+1=4*（an+1）/an+2两边取倒数1/（a(n+1)+1）=（an+2）/4*（an+1） =（an+1+1）/4*（an+1） =1/4+(1/4)*(1/an+1)∵bn=1/(an+1)∴b(n+1）=1/4+(1/4)*bn上式可化为b(n+1）-1/3=(1/4)*（bn-1/3）所以数列{bn-1/3}是公比为1/4的等比数列bn-1/3=(b1-1/3)*(1/4)^(n-1) =(1/(0.5+1)-1/3)*(1/4)^(n-1) =(1/3)*(1/4)^(n-1)bn=(1/3)*(1/4)^(n-1)+1/3