Math Problem Statement

Work out the value of Fn−1Fn+1 −Fn2 for n = 1,2,3,4,5. You should see a pattern; prove that this pattern always holds.

Solution

To solve the expression Fn1Fn+1Fn2F_{n-1}F_{n+1} - F_n^2 for n=1,2,3,4,5n = 1, 2, 3, 4, 5 where FnF_n represents the Fibonacci sequence, let's first recall the Fibonacci sequence:

F1=1,F2=1,F3=2,F4=3,F5=5,F6=8,and so on.F_1 = 1, \quad F_2 = 1, \quad F_3 = 2, \quad F_4 = 3, \quad F_5 = 5, \quad F_6 = 8, \quad \text{and so on.}

The expression we want to calculate is Fn1Fn+1Fn2F_{n-1}F_{n+1} - F_n^2.

Step 1: Calculate the expression for specific values of nn

  1. For n=1n = 1: F0F2F12=0112=01=1F_0F_2 - F_1^2 = 0 \cdot 1 - 1^2 = 0 - 1 = -1

  2. For n=2n = 2: F1F3F22=1212=21=1F_1F_3 - F_2^2 = 1 \cdot 2 - 1^2 = 2 - 1 = 1

  3. For n=3n = 3: F2F4F32=1322=34=1F_2F_4 - F_3^2 = 1 \cdot 3 - 2^2 = 3 - 4 = -1

  4. For n=4n = 4: F3F5F42=2532=109=1F_3F_5 - F_4^2 = 2 \cdot 5 - 3^2 = 10 - 9 = 1

  5. For n=5n = 5: F4F6F52=3852=2425=1F_4F_6 - F_5^2 = 3 \cdot 8 - 5^2 = 24 - 25 = -1

Step 2: Identify the pattern

The results alternate between 1-1 and 11 for successive values of nn:

  • n=1n = 1: 1-1
  • n=2n = 2: 11
  • n=3n = 3: 1-1
  • n=4n = 4: 11
  • n=5n = 5: 1-1

Step 3: Prove the pattern

Let's prove that Fn1Fn+1Fn2=(1)nF_{n-1}F_{n+1} - F_n^2 = (-1)^n for all n1n \geq 1.

Proof by induction:

Base case: n=1n = 1

For n=1n = 1: F0F2F12=0112=1F_0F_2 - F_1^2 = 0 \cdot 1 - 1^2 = -1 which is equal to (1)1=1(-1)^1 = -1.

Inductive step:

Assume that the formula holds for n=kn = k, i.e., Fk1Fk+1Fk2=(1)kF_{k-1}F_{k+1} - F_k^2 = (-1)^k.

We need to show that it holds for n=k+1n = k+1, i.e., FkFk+2Fk+12=(1)k+1F_kF_{k+2} - F_{k+1}^2 = (-1)^{k+1}.

Using the Fibonacci recurrence relation Fk+2=Fk+1+FkF_{k+2} = F_{k+1} + F_k, we can write: FkFk+2=Fk(Fk+1+Fk)=FkFk+1+Fk2F_kF_{k+2} = F_k(F_{k+1} + F_k) = F_kF_{k+1} + F_k^2 Thus, FkFk+2Fk+12=(FkFk+1+Fk2)Fk+12=FkFk+1Fk+12+Fk2F_kF_{k+2} - F_{k+1}^2 = (F_kF_{k+1} + F_k^2) - F_{k+1}^2 = F_kF_{k+1} - F_{k+1}^2 + F_k^2

From the inductive hypothesis, Fk1Fk+1Fk2=(1)kF_{k-1}F_{k+1} - F_k^2 = (-1)^k, hence: FkFk+2Fk+12=(1)kF_kF_{k+2} - F_{k+1}^2 = (-1)^k

Therefore, the formula holds for n=k+1n = k+1 if it holds for n=kn = k. By the principle of mathematical induction, the formula Fn1Fn+1Fn2=(1)nF_{n-1}F_{n+1} - F_n^2 = (-1)^n holds for all n1n \geq 1.

Summary

The value of Fn1Fn+1Fn2F_{n-1}F_{n+1} - F_n^2 alternates between 1-1 and 11 depending on whether nn is odd or even, and it is equal to (1)n(-1)^n.

Would you like to explore any further details or have any specific questions?

Related Questions

  1. How can the Fibonacci sequence be generalized to other recurrence relations?
  2. What is the significance of the identity Fn1Fn+1Fn2=(1)nF_{n-1}F_{n+1} - F_n^2 = (-1)^n in combinatorics?
  3. Can similar identities be derived for other sequences like the Lucas sequence?
  4. How is the Fibonacci sequence connected to the golden ratio?
  5. What other mathematical properties does the Fibonacci sequence exhibit?

Tip:

The identity Fn1Fn+1Fn2=(1)nF_{n-1}F_{n+1} - F_n^2 = (-1)^n is an example of a Cassini identity, named after the astronomer Giovanni Cassini.

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Math Problem Analysis

Mathematical Concepts

Fibonacci Sequence
Recurrence Relations
Mathematical Induction

Formulas

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Theorems

Cassini Identity

Suitable Grade Level

Advanced High School