摘要：若数列{an}前8项的值各异且an+8=an对任意n∈N*都成立.则下列数列中可取遍{an}前8项值的数列为( ) A.{a2k+1} B.{a3k+1} C.{a4k+1} D.{a6k+1} [答案]B [解析]∵2k+1,4k+1,6k+1只能取奇数.又周期为8.故排除A.C.D.选B.
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若数列{an}前8项的值各异且an+8=an对任意n∈N*都成立，则下列数列中可取遍{an}前8项值的数列为( )
A.{a2k+1} B.{a3k+1} C.{a4k+1} D.{a6k+1}
查看习题详情和答案>>若数列{an}前8项的值各异且an+8＝an对任意的n∈N*都成立．则下列数列中可取遍{an}前8项值的数列为
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A．{a2k+1}
B．{a3k+1}
C．{a4k+1}
D．{a6k+1}