Question

设a1=2,a(n+1)=2/(an+1),bn=|(an+2)/(an-1)|-1,n∈N*,则b2011=?

Answer 1

显然an都是正有理数，令an=p/q，p、q∈N⁺，则a(n+1)=2/(p/q+1)=2q/(p+q)
bn=|(an+2)/(an-1)|-1=|(p/q+2)/(p/q-1)|-1=|(p+2q)/(p-q)|-1，即bn+1=|(p+2q)/(p-q)|
b(n+1)=|(a(n+1)+2)/(a(n+1)-1)|-1=|(2q/(p+q)+2)/(2q/(p+q)-1)|-1
=2|(p+2q)/(p-q)|-1
即b(n+1)+1=2|(p+2q)/(p-q)|=2(bn+1)，∴bn+1是以2为公比的等比数列。

由a₁=2可知b₁=3，b₁+1=4
于是b(n)+1=(b₁+1)·2ⁿ⁻¹=2ⁿ⁺¹，b(n)=2ⁿ⁺¹-1
∴b₂₀₁₁=2²⁰¹²-1