a(n+1)·√an=8a1=1>0,假设当n=k(k∈N+)时,ak>0,则a(k+1)=8/√ak>0k为任意正整数,因此,对于任意正整数n,an恒>0log2[a(n+1)√an]=log2(8)log2[a(n+1)]+(1/2)log2(an)=3log2[a(n+1)] -2=-(1/2)log2(an) +1=(-1/2)[log2...谢了，是不是an=2^[2+(-1/2)^(n-2)]中间是+号是的。