解:(1)由题意得a2,a4,a6,a8,…成等比数列,且公比q=()=, 所以a2n=()n-4. (2)由数列{an}是“J4型”数列,得 a1,a5,a9,a13,a17,a21,…成等比数列,设公比为t. 由数列{an}是“J3型”数列,得 a1,a4,a7,a10,a13,…成等比数列,设公比为α1; a2,a5,a8,a11,a14,…成等比数列,设公比为α2; a3,a6,a9,a12,a15,…成等比数列,设公比为α3. 则=α14=t3,=α24=t3,=α34=t3. 所以α1=α2=α3,不妨记α=α1=α2=α3,且t=α. 于是a3k-2=a1αk-1=a1()(3k-2)-1, a3k-1=a5αk-2=a1tαk-2=a1αk-=a1()(3k-1)-1, a3k=a9αk-3=a1t2αk-3=a1αk-=a1()3k-1, 所以an=a1()n-1,故{an}为等比数列. |