Exercise 4.4.6 ($F_{n+1} F_{n-1} - F_n^2 = (-1)^n$)

Question

Prove that $F_{n+1} F_{n-1} - F_n^2 = (-1)^n$.

Richard Ganaye · Accepted Answer

ewcommand{\N}{\mathbb{N}}

{\bf First proof.}
\begin{proof} 
Put 
$$\lambda = \frac{1- \sqrt{5}}{2},\qquad \mu =  \frac{1+ \sqrt{5}}{2},$$
the roots of the polynomial $x^2 - x - 1$. Then $\lambda + \mu = 1, \ \lambda \mu = -1$.

By (4.6),
\begin{align*}
F_{n+1} F_{n-1} - F_n^2 &= \frac{\mu^{n+1} - \lambda^{n+1}}{\mu - \lambda} \, \frac{\mu^{n-1} - \lambda^{n-1}}{\mu - \lambda} - \left( \frac{\mu^{n} - \lambda^{n}}{\mu - \lambda}ight)^2\
&=\frac{1}{(\mu - \lambda)^2} \left[ (\mu^{n+1} - \lambda^{n+1})(\mu^{n-1} - \lambda^{n-1}) - (\mu^{n} - \lambda^{n})^2ight]\
&= \frac{1}{(\mu - \lambda)^2}  ( 2 \lambda^n \mu^n - \lambda^{n+1} \mu^{n-1} - \lambda^{n-1} \mu^{n+1})\
&=\frac{(\lambda \mu)^{n-1}}{(\mu - \lambda)^2} (2 \lambda \mu - \lambda^2 - \mu^2)\
&= -(\lambda \mu)^{n-1}\
& = (-1)^n.
\end{align*}

\end{proof}

\bigskip
{\bf Second proof.} 
\begin{proof}
Put $A = \begin{pmatrix} 0 & 1 \ 1 & 1\end{pmatrix}$, and $X_n = \begin{pmatrix} F_n\ F_{n+1}\end{pmatrix}$.
Then, for all $n\in \N$,
$$X_{n+1} = \begin{pmatrix} F_{n+1}\ F_{n+2}\end{pmatrix} =  \begin{pmatrix} 0 & 1 \ 1 & 1\end{pmatrix} \begin{pmatrix} F_n\ F_{n+1}\end{pmatrix} = A X_n.$$
Thus
$$ \begin{pmatrix} F_n & F_{n+1} \ F_{n+1}& F_{n+2}\end{pmatrix}  = \begin{pmatrix} 0 & 1 \ 1 & 1\end{pmatrix} \begin{pmatrix} F_{n-1} & F_{n} \ F_{n}& F_{n+1}\end{pmatrix} .$$
We define $V_n =  \begin{pmatrix} F_{n-1} & F_{n} \ F_{n}& F_{n+1}\end{pmatrix} $ for all $n \in \N^*$. Then $V_{n+1} = A V_{n}$, so $V_n = A^{n-1} V_1$ for all $n \geq 1$, where
$$V_1  =\begin{pmatrix} F_0 & F_1 \ F_1 & F_2\end{pmatrix} = \begin{pmatrix} 0 & 1 \ 1 & 1\end{pmatrix} = A.$$
Therefore, for all $n \in \N$, $V_n = A^n$, that is
$$V_n = \  \begin{pmatrix} F_{n-1} & F_{n} \ F_{n}& F_{n+1}\end{pmatrix} = \begin{pmatrix} 0 & 1 \ 1 & 1\end{pmatrix}^n = A^n.$$

Taking the determinants, we obtain $\det(V_n) = \det(A)^n$, where $\det(A) = -1$, so for all positive integers $n$,
$$F_{n+1} F_{n-1} - {F_{n}}^2 = (-1)^n.$$
\end{proof}

Exercise 4.4.6 ($F_{n+1} F_{n-1} - F_n^2 = (-1)^n$)

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Exercise 4.4.6 ($F_{n+1} F_{n-1} - F_n^2 = (-1)^n$)

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