text: # My Favorite One-Liners: Part 28
  
  In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Today’s quip is one that I’ll use when simple techniques get used in a complicated way.
  
  Consider the solution of the linear recurrence relation
  
  $Q_n = Q_{n-1} + 2 Q_{n-2}$,
  
  where $F_0 = 1$ and $F_1 = 1$. With no modesty, I call this one the Quintanilla sequence when I teach my students — the forgotten little brother of the Fibonacci sequence.
  
  To find the solution of this linear recurrence relation, the standard technique — which is a pretty long procedure — is to first solve the characteristic equation, from $Q_n - Q_{n-1} - 2 Q_{n-2} = 0$, we obtain the characteristic equation
  
  $r^2 - r - 2 = 0$
  
  This can be solved by any standard technique at a student’s disposal. If necessary, the quadratic equation can be used. However, for this one, the left-hand side simply factors:
  
  $(r-2)(r+1) = 0$
  
  $r=2 \qquad \hbox{or} \qquad r = -1$
  
  (Indeed, I “developed” the Quintanilla equation on purpose, for pedagogical reasons, because its characteristic equation has two fairly simple roots — unlike the characteristic equation for the Fibonacci sequence.)
  
  From these two roots, we can write down the general solution for the linear recurrence relation:
  
  $Q_n = \alpha_1 \times 2^n + \alpha_2 \times (-1)^n$,
  
  where $\alpha_1$ and $\alpha_2$ are constants to be determined. To find these constants, we plug in $n =0$:
  
  $Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0$.
  
  To find these constants, we plug in $n =0$:
  
  $Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0$.
  
  We then plug in $n =1$:
  
  $Q_1 = \alpha_1 \times 2^1 + \alpha_2 \times (-1)^1$.
  
  Using the initial conditions gives
  
  $1 = \alpha_1 + \alpha_2$
  
  $1 = 2 \alpha_1 - \alpha_2$
  
  This is a system of two equations in two unknowns, which can then be solved using any standard technique at the student’s disposal. Students should quickly find that $\alpha_1 = 2/3$ and $\alpha_2 = 1/3$, so that
  
  $Q_n = \displaystyle \frac{2}{3} \times 2^n + \frac{1}{3} \times (-1)^n = \frac{2^{n+1} + (-1)^n}{3}$,
  
  Although this is a long procedure, the key steps are actually first taught in Algebra I: solving a quadratic equation and solving a system of two linear equations in two unknowns. So here’s my one-liner to describe this procedure:
  
  This is just an algebra problem on steroids.
  
  Yes, it’s only high school algebra, but used in a creative way that isn’t ordinarily taught when students first learn algebra.
  
  I’ll use this “on steroids” line in any class when a simple technique is used in an unusual — and usually laborious — way to solve a new problem at the post-secondary level.
explanation: The document describes a method for solving a linear recurrence relation, specifically the "Quintanilla sequence," which is similar to the Fibonacci sequence. It outlines the steps to solve the characteristic equation, find the general solution, and determine the constants using initial conditions. The text includes a detailed explanation of the process and concludes with a one-liner to emphasize the complexity of the problem despite using basic algebraic techniques.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # My Favorite One-Liners: Part 28
  
  In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Today’s quip is one that I’ll use when simple techniques get used in a complicated way.
  
  Consider the solution of the linear recurrence relation
  
  $Q_n = Q_{n-1} + 2 Q_{n-2}$,
  
  where $F_0 = 1$ and $F_1 = 1$. With no modesty, I call this one the Quintanilla sequence when I teach my students — the forgotten little brother of the Fibonacci sequence.
  
  To find the solution of this linear recurrence relation, the standard technique — which is a pretty long procedure — is to first solve the characteristic equation, from $Q_n - Q_{n-1} - 2 Q_{n-2} = 0$, we obtain the characteristic equation
  
  $r^2 - r - 2 = 0$
  
  This can be solved by any standard technique at a student’s disposal. If necessary, the quadratic equation can be used. However, for this one, the left-hand side simply factors:
  
  $(r-2)(r+1) = 0$
  
  $r=2 \qquad \hbox{or} \qquad r = -1$
  
  (Indeed, I “developed” the Quintanilla equation on purpose, for pedagogical reasons, because its characteristic equation has two fairly simple roots — unlike the characteristic equation for the Fibonacci sequence.)
  
  From these two roots, we can write down the general solution for the linear recurrence relation:
  
  $Q_n = \alpha_1 \times 2^n + \alpha_2 \times (-1)^n$,
  
  where $\alpha_1$ and $\alpha_2$ are constants to be determined. To find these constants, we plug in $n =0$:
  
  $Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0$.
  
  To find these constants, we plug in $n =0$:
  
  $Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0$.
  
  We then plug in $n =1$:
  
  $Q_1 = \alpha_1 \times 2^1 + \alpha_2 \times (-1)^1$.
  
  Using the initial conditions gives
  
  $1 = \alpha_1 + \alpha_2$
  
  $1 = 2 \alpha_1 - \alpha_2$
  
  This is a system of two equations in two unknowns, which can then be solved using any standard technique at the student’s disposal. Students should quickly find that $\alpha_1 = 2/3$ and $\alpha_2 = 1/3$, so that
  
  $Q_n = \displaystyle \frac{2}{3} \times 2^n + \frac{1}{3} \times (-1)^n = \frac{2^{n+1} + (-1)^n}{3}$,
  
  Although this is a long procedure, the key steps are actually first taught in Algebra I: solving a quadratic equation and solving a system of two linear equations in two unknowns. So here’s my one-liner to describe this procedure:
  
  This is just an algebra problem on steroids.
  
  Yes, it’s only high school algebra, but used in a creative way that isn’t ordinarily taught when students first learn algebra.
  
  I’ll use this “on steroids” line in any class when a simple technique is used in an unusual — and usually laborious — way to solve a new problem at the post-secondary level.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document describes a method for solving a linear recurrence relation, specifically the "Quintanilla sequence," which is similar to the Fibonacci sequence. It outlines the steps to solve the characteristic equation, find the general solution, and determine the constants using initial conditions. The text includes a detailed explanation of the process and concludes with a one-liner to emphasize the complexity of the problem despite using basic algebraic techniques.
  
  Label: no-question