已知等比数列{a
n},a
2>a
3=1,则使不等式(a
1-
)+(a
2-
)+…+(a
n-
)≥0成立的最大自然数n是( )
A. 4
B. 5
C. 6
D. 7
人气:154 ℃ 时间:2019-08-20 04:57:34
解答
设公比为q,a2>a3=1,则有1>q>0可知n>3时,有an-1an<0a3=a1q2=1得a1=1q2则有a5=a1q4=q2=1a1,同理有a2=1a4得(a1-1a1)+(a2-1a2)+(a3-1a3)+(a4-1a4)+(a5-1a5)=0∴不等式(a1-1a1)+(a2-1a2)+…+(an-...
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