已知函数f（x）=2x次方-2负x次方数列｛an｝满足f（log2an）=-2n.（1） 证明数列｛an｝递减数列

an>0f(x) = 2^x - 2^(-x) = [2^(2x) - 1] / 2^xf(㏒2 an) = 2^(㏒2 an) - 2^(-㏒2 an) = an - 1/an = -2n同样, f(㏒2 an+1) = (an+1) - 1/(an+1) = -2(n+1)以上两式相减(an+1) - an + 1/an - 1/(an+1) = -2[(an+1) - an] (1 + 1/(an* an+1)^) =-2