已知数列{a
n},若a
1=14,
an+1=an−(n∈N
*),则使a
n•a
n+2<0成立的n的值是______.
人气:488 ℃ 时间:2020-09-11 08:57:18
解答
∵a1=14,an+1=an−23(n∈N*),∴数列{an}是首项为14,公差为-23的等差数列,∴an=14+(n−1)×(−23)=-23n+443,∴an•an+2=(-23n+443)[-23(n+2)+443]=49n2−1689n+17609,∵an•an+2<0,∴49n2−1689n+17609...
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