(理)数列{an},若对任意的k∈N*,满足
(1)若数列{an}为“跳跃等比数列”,则满足bk=a2k•a2k-1(k∈N*)的数列{bn}是等比数列; (2)若数列{an}为“跳跃等比数列”,则满足bk=
(3)若数列{an}为等比数列,则数列{(-1)nan}是“跳跃等比数列”; (4)若数列{an}为等比数列,则满足bn=
(5)若数列{an}和{bn}都是“跳跃等比数列”,则数列{an•bn}也是“跳跃等比数列”;其中正确的命题个数为( )
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(1)若数列{an}为“跳跃等比数列”,则
=bk+1 bk
=q2•q1(常数),根据等比数列的定义可知数列{bn}是等比数列,故正确;a2k+2• a2k+1 a2k•a2k-1
(2)若数列{an}为“跳跃等比数列”,则
=bk+1 bk
=a2k+2•a2k-1 a2k•a2k+1
(常数),根据等比数列的定义可知数列{bn}是等比数列,故正确;q2 q1
(3)若数列{an}为等比数列,假设公比为q,则
=q2,-a2k+1 -a2k-1
=q2,公比相等不符合定义,∴数列{(-1)nan}不是“跳跃等比数列”,故不正确;a2k+2 a2k
(4)若数列{an}为等比数列,假设公比为q,假设n=2k-1,则
=bn+1 bn
=q q a 2k
≠常数,故数列{bn}不是“跳跃等比数列”,故不正确;1 a 2k
(5)若数列{an}和{bn}都是“跳跃等比数列”,则
=q1,a2k+1 a2k-1
=q2a2k+2 a2k
是常数且不相等), &(q1,q2
= p1,b2k+1 b2k-1
=p2(p1,p2是常数且不相等),那么数列{an•bn}也是“跳跃等比数列”,故正确.b2k+2 b2k
故选C.