Day: March 27, 2023

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202303270928 Problem 1.1.1

Let F_1, F_2, F_3, … be the Fibonacci sequence. Prove that

F_{n+1}F_{n+2}-F_{n}F_{n+3}=(-1)^n

for all positive integers n.

Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.

Roughwork.

Let P(n) be the statement to prove. When n=1,

P(1):\quad F_{2}F_{3}-F_{1}F_{4}\stackrel{\textrm{?}}{=}-1

is to check.

\begin{aligned} F_1 & = 1\\ F_2 & = 1\\ F_3 & = 2\\ F_4 & = 3\\ \dots & \dots \\ F_{n} & = F_{n-1} + F_{n-2}\quad\textrm{for integers }n\geqslant 3 \\ \end{aligned}

First, P(1) is true.

Next, be P(n) true. WTS P(n+1) true whenever P(n) true:

\begin{aligned} P(n+1): & \quad F_{(n+1)+1}F_{(n+1)+2}-F_{n+1}F_{(n+1)+3}\stackrel{\textrm{?}}{=}(-1)^{n+1} \\ \textrm{LHS} & = F_{n+2}F_{n+3}-F_{n+1}F_{n+4} \\ & = (F_{n+1}+F_n)F_{n+3}-F_{n+1}(F_{n+3}+F_{n+2})\\ & = F_{n}F_{n+3}-F_{n+1}F_{n+2} \\ & = -(F_{n+1}F_{n+2}-F_{n}F_{n+3}) \\ & \stackrel{.}{=} -((-1)^n) \\ & = (-1)^{n+1} \\ & = \textrm{RHS} \\ \end{aligned}

Lastly, as P(1) true and P(n)\Rightarrow P(n+1), by principle of mathematical induction, the statement is proven for all n\in\mathbb{Z}^+.