The problem is:
$$F_n \leqslant 2F_{n-1}\quad\text{for every integer} \quad n \geqslant 2.$$
I got the smallest case, I just don't know how to get the assumption and the rest of it
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1Remember that $F_n = F_{n-1} + F_{n-2}$. What relation between $F_{n-1}$ and $F_{n-2}$ gives you the desired conclusion? – Daniel Fischer Sep 29 '14 at 14:40
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After i do the smallest case which is n = 2, could I get help with the assumption? is it just Fk <= 2Fk-1 ??? – Jacolby Hartson Sep 29 '14 at 14:44
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You use complete induction: assume that $F_m \leqslant 2F_{m-1}$ for all $m<n$ and prove it for $n$:
$F_n = F_{n-1}+F_{n-2} \le 2F_{n-2}+2F_{n-3} = 2(F_{n-2}+F_{n-3}) = 2F_{n-1}$
You need to start by proving it for $n=2$ and $n=3$.
lhf
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Yes, you assume that $F_k \lt 2F_{k-1}$ and want to prove that $F_{k+1} \lt 2F_k$ So write $F_{k+1}= $ (what? you only know one thing about it) then apply what you know about the numbers to put an upper bound on this. You do need to know the Fibonacci numbers are positive. You might look at this answer
Ross Millikan
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Knowing that the Fibonacci numbers are non-negative tells you that the sequence is non-decreasing, so $F_{k + 1} = F_k + F_{k - 1} \leq F_k + F_k = 2F_k$. – N. F. Taussig Sep 29 '14 at 15:30
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@N.F.Taussig, that's one way but I had something different in mind. See my answer. – lhf Sep 29 '14 at 15:53
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