摘要:②BN==1/2
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(理)数列{an},若对任意的k∈N*,满足
=q1,
=q2
是常数且不相等),则称数列{an}为“跳跃等比数列”,则下列关于“跳跃等比数列”的命题:
(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=
(k∈N*)的数列{bn}是等比数列;
(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=
(k∈N*)的数列{bn}是“跳跃等比数列”;
(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( )
a2k+1 |
a2k-1 |
a2k+2 |
a2k |
|
(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=
a2k |
a2k-1 |
(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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