Question

若数列{a_n}的前8项值各异,且a_n+8=a_n,对任意的n∈N^*都成立,则下列数列中可取遍{a_n}的前8项值的数列为(    )

A.{a_2k+1}           B.{a_3k+1}            C.{a_4k+1}          D.{a_6k+1}

Answer 1

解析:对于B,k=0时,得a₁;k=1时,得a₄;

k=2时,得a₇;k=3时,得a₁₀=a₂;

k=4时,得a₁₃=a₅;

k=5时,得a₁₆=a₈;k=6时,得a₁₉=a₃;

k=7时,得a₂₂=a₆,….选B.

答案:B