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(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.
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.