Question

已知a1=1,a2=3,a(n+1)=1+2/an,求an

Answer 1

a(n+1)=1+2/an,
a(n+1)-2=1+2/an-2,
a(n+1)-2=2/an-1,
a(n+1)-2=(2-an)/an,
取倒数得：1/( a(n+1)-2)= an / (2-an),
1/( a(n+1)-2)= [an-2+2] / (2-an),
1/( a(n+1)-2)=-1+2/ (2-an),
1/( a(n+1)-2)=-1-2/ (an -2),
设1/ (an -2)=bn,
上式化为：b(n+1)= -1-2 bn,
b(n+1)+1/3=-2(bn+1/3),
数列{ bn+1/3}是等比数列，公比是-2，首项是b1+1/3=-2/3,
所以bn+1/3=-2/3•(-2)^(n-1),
即1/ (an -2) +1/3=-2/3•(-2)^(n-1),
∴an=3/[(-2)^n-1]+2.