Let's begin by calculating the expression Fn−1Fn+1−Fn2 for the first few values of n, where Fn denotes the n-th Fibonacci number.
Step 1: Recall the Fibonacci sequence
The Fibonacci sequence is defined as:
F1=1,F2=1
Fn=Fn−1+Fn−2for n≥3
So, the first few Fibonacci numbers are:
F1=1,F2=1,F3=2,F4=3,F5=5,F6=8,…
Step 2: Calculate the expression for n=1,2,3,4,5
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For n=1:
F0=0,F2=1,F1=1
F0F2−F12=0×1−12=0−1=−1
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For n=2:
F1=1,F3=2,F2=1
F1F3−F22=1×2−12=2−1=1
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For n=3:
F2=1,F4=3,F3=2
F2F4−F32=1×3−22=3−4=−1
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For n=4:
F3=2,F5=5,F4=3
F3F5−F42=2×5−32=10−9=1
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For n=5:
F4=3,F6=8,F5=5
F4F6−F52=3×8−52=24−25=−1
Step 3: Observe the pattern
From the calculations above, we notice the following pattern:
- When n is odd, Fn−1Fn+1−Fn2=−1.
- When n is even, Fn−1Fn+1−Fn2=1.
Step 4: Prove that this pattern always holds
We need to prove that for any n≥1:
Fn−1Fn+1−Fn2=(−1)n
Proof by Induction
Base Case:
For n=1, we've already calculated:
F0F2−F12=0×1−12=−1
which is (−1)1.
Inductive Step:
Assume that for some n=k, the pattern holds:
Fk−1Fk+1−Fk2=(−1)k
We need to show that it holds for n=k+1, i.e., we need to prove:
FkFk+2−Fk+12=(−1)k+1
Using the Fibonacci identity Fk+2=Fk+1+Fk, we can rewrite the left side:
FkFk+2−Fk+12=Fk(Fk+1+Fk)−Fk+12
=FkFk+1+Fk2−Fk+12
=(FkFk+1−Fk+12)+Fk2
=Fk+1(Fk−Fk+1)+Fk2
But since Fk−Fk+1=−Fk−1, this simplifies to:
=−Fk+1Fk−1+Fk2
=−(Fk−1Fk+1−Fk2)
By the inductive hypothesis:
=−((−1)k)=(−1)k+1
Thus, the pattern Fn−1Fn+1−Fn2=(−1)n holds for all n≥1.
Would you like to explore more about this proof or Fibonacci identities?
Related Questions:
- How can you prove other identities involving Fibonacci numbers?
- What is the significance of the alternating pattern in Fibonacci-related expressions?
- Can this pattern be generalized to other recursive sequences?
- How does the Fibonacci sequence relate to the golden ratio?
- What are the applications of Fibonacci identities in real-world problems?
Tip:
When dealing with sequences like Fibonacci, always try to use induction or known identities to simplify complex expressions.