Question

高一数学 谢谢~

Answer 1

a(n+1) = 2/(an+1)
a(n+1) -1 = (1-an)/(an +1)
1/[a(n+1) -1] = -(an +1)/(an -1)
=-( 1+ 2/(an-1) )
1/[a(n+1) -1]+1/3 = -2( 1/(an-1) + 1/3 )
[1/[a(n+1) -1]+1/3]/( 1/(an-1) + 1/3 ) = -2
( 1/(an-1) + 1/3 )/( 1/(a1-1) + 1/3 ) = (-2)^(n-1)
1/(an-1) + 1/3 = -(2/3) (-2)^n
1/(an -1) = -[1/3+ (2/3)(-2)^n ]
an -1 = -1/[1/3+ (2/3)(-2)^n ]
an = 1- 1/[1/3+ (2/3)(-2)^n ]

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

追问

答案能用汉语说吗？