SOLUTION: Show that whenever n ≥ 3, fn > ßn-2 , where ß = (1 + √ 5)∕2.
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Question 1193450: Show that whenever n ≥ 3, fn > ßn-2 , where ß = (1 + √ 5)∕2.
Answer by ikleyn(52799) (Show Source): You can put this solution on YOUR website!
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Show that whenever n ≥ 3, fn > ßn-2 , where ß = (1 + √ 5)∕2.
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This problem was posted approximately a month or two ago in the same form,
and I explained/responded, that in this form it is DEFECTIVE and can not be solved.
See the link
https://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.1192804.html
This time, I confirm AGAIN that previous DIAGNOSIS.
Some "visitors" at this forum are so SLOW that one month is not enough time for them to get the meaning of my message.
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