Fibonacci Sequence - Induction.

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SUMMARY

The discussion focuses on proving the inequality Fn ≤ (7/4)n for all n, where Fn represents the Fibonacci sequence defined by Fn = Fn-1 + Fn-2. The proof utilizes mathematical induction, starting with the assumption that P(k) holds true for some k. The participants explore the transition from Fk to Fk+1, ultimately leading to the expression (7/4)k+1. A critical simplification step involves recognizing that (7/4)k-1 + (7/4)k can be expressed in terms of (7/4)k, which is essential for completing the proof.

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Prove Fn ≤ (7/4)n for all n, 0≤n

Fn = Fn-1 + Fn-2

Let P(n) be true for some n = k, for 0≤k

Let n = k+1

Fk+1 ≤ (7/4)k+1

LHS: Fk+1 = Fk + Fk-1 ≤ Fk-1 + (7/4)k ≤ (7/4)k-1 + (7/4)k

This last line is where I'm stuck, I feel like either I messed up early on, or I'm missing a way of simplifying this to look like (7/4)k+1
 
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glover_m said:
Prove Fn ≤ (7/4)n for all n, 0≤n

Fn = Fn-1 + Fn-2

Let P(n) be true for some n = k, for 0≤k

Let n = k+1

Fk+1 ≤ (7/4)k+1

LHS: Fk+1 = Fk + Fk-1 ≤ Fk-1 + (7/4)k ≤ (7/4)k-1 + (7/4)k

This last line is where I'm stuck, I feel like either I messed up early on, or I'm missing a way of simplifying this to look like (7/4)k+1
Well, [itex](7/4)^{k-1}= (4/7)(7/4)^k[/itex] so [itex](7/4)^{k-1}+ (7/4)^k= (4/7+ 1)(7/4)^k[/itex]. And 4/7+ 1= 11/7= (11/7)(4/7)(7/4)= (44/49)(7/4) and 44/49< 1.
 

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