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.