f(log2an)=2^(log2an)-2^(-log2an)=an-1/an=-2n
=>an^2+2n*an-1=0
因为log2an有意义
所以an>0
所以an=√(n^2+1) -n
an =√(n^2+1) -n =1/[√(n^2+1) +n]
于是显然有a(n+1)