Question

已知数列an满足a1=2,an=（a（n-1）+2）/（2a（n-1）+1）,求通项公式an

Answer 1

an = (a(n-1)+2)/(2a(n-1)+1)

an -1 =(a(n-1)+2)/(2a(n-1)+1) -1

= -(a(n-1)-1)/(2a(n-1)+1)

1/(an -1) = -(2a(n-1)+1)/(a(n-1)-1)

= -2 - 3/(a(n-1) -1)

1/(an -1) + 1/2 = -3[1/(a(n-1) -1) +1/2]

{1/(an -1) + 1/2}是等比数列, q=-3

1/(an -1) + 1/2 = (-3)^(n-1) .(1/(a1 -1) + 1/2)

=-(1/2)(-3)^n

an -1 = -2/[1 + (-3)^n ]

an = 1 - 2/[1 + (-3)^n ]