# Tag Archives: algebra

## Powers of 2

Yesterday, James Tanton posted a fun little problem on Twitter:

So, 2 is one more than $1=2^0$, and 8 is one less than 9=2^3$, and Dr. Tanton wants to know if there are any other powers of two that are within one unit of a perfect square. While this problem may not have any “real-life relevance”, it demonstrates what I describe as the power and creativity of mathematics. Among the infinite number of powers of two, how can someone know for certain if any others are or are not within one unit of a perfect square? No one will ever be able to see every number in the list of powers of two, but variables and mathematics give you the tools to deal with all possibilities at once. For this problem, let D and N be positive integers. Translated into mathematical language, Dr. Tanton’s problem is equivalent to asking if there are values of D and N for which $2^D=N^2 \pm 1$. With a single equation in two unknowns, this is where observation and creativity come into play. I suspect there may be more than one way to approach this, but my solution follows. Don’t read any further if you want to solve this for yourself. WARNING: SOLUTION ALERT! Because D and N are positive integers, the left side of $2^D=N^2 \pm 1$, is always even. That means $N^2$, and therefore N must be odd. Because N is odd, I know $N=2k+1$ for some whole number k. Rewriting our equation gives $2^D=(2k+1)^2 \pm 1$, and the right side equals either $4k^2+4k$ or $4k^2+4k+2$. Factoring the first expression gives $2^D=4k^2+4K=4k(k+1)$. Notice that this is the product of two consecutive integers, k and $k+1$, and therefore one of these factors (even though I don’t know which one) must be an odd number. The only odd number that is a factor of a power of two is 1, so either $k=1$ or $k+1=1 \rightarrow k=0$. Now, $k=1 \longrightarrow N=3 \longrightarrow D=3$ and $k=0 \longrightarrow N=1 \longrightarrow D=0$, the two solutions Dr. Tanton gave. No other possibilities are possible from this expression, no matter how far down the list of powers of two you want to go. But what about the other expression? Factoring again gives $2^D=4k^2+4k+2=2 \cdot \left( 2k^2+2k+1 \right)$. The expression in parentheses must be odd because its first two terms are both multiplied by 2 (making them even) and then one is added (making the overall sum odd). Again, 1 is the only odd factor of a power of two, and this happens in this case only when $k=0 \longrightarrow N=1 \longrightarrow D=0$, repeating a solution from above. Because no other algebraic solutions are possible, the two solutions Dr. Tanton gave in the problem statement are the only two times in the entire universe of perfect squares and powers of two where elements of those two lists are within a single unit of each other. Math is sweet. ## Dynamic Linear Programming My department is exploring the pros and cons of different technologies for use in teaching our classes. Two teachers shared ways to use Desmos and GeoGebra in lessons using inequalities on one day; we explored the same situation using the TI-Nspire in the following week’s meeting. For this post, I’m assuming you are familiar with solving linear programming problems. Some very nice technology-assisted exploration ideas are developed in the latter half of this post. My goal is to show some cool ways we discovered to use technology to evaluate these types of problems and enhance student exploration. Our insights follow the section considering two different approaches to graphing the feasible region. For context, we used a dirt-biker linear programming problem from NCTM’s Illuminations Web Pages. Assuming x = the number of Riders built and = the number of Rovers built, inequalities for this problem are We also learn on page 7 of the Illuminations activity that Apu makes a$15 profit on each Rider and $30 per Rover. That means an Optimization Equation for the problem is $Profit=15x+30y$. GRAPHING THE FEASIBLE REGION: Graphing all of the inequalities simultaneously determines the feasible region for the problem. This can be done easily with all three technologies, but the Nspire requires solving the inequalities for y first. Therefore, the remainder of this post compares the Desmos and GeoGebra solutions. Because the Desmos solutions are easily accessible as Web pages and not separate files, further images will be from Desmos until the point where GeoGebra operates differently. Both Desmos and GeoGebra can graph these inequalities from natural inputs–inputing math sentences as you would write them from the problem information: without solving for a specific variable. As with many more complicated linear programming problems, graphing all the constraints at once sometimes makes a visually complicated feasible region graph. So, we decided to reverse all of our inequalities, effectively shading the non-feasible region instead. Any points that emerged unshaded were possible solutions to the Dirt Bike problem (image below, file here). All three softwares shift properly between solid and dashed lines to show respective included and excluded boundaries. Traditional Approach – I (as well as almost all teachers, I suspect) have traditionally done some hand-waving at this point to convince (or tell) students that while any ordered pair in the unshaded region or on its boundary (all are dashed) is a potential solution, any optimal solution occurs on the boundary of the feasible region. Hopefully teachers ask students to plug ordered pairs from the feasible region into the Optimization Equation to show that the profit does vary depending on what is built (duh), and we hope they eventually discover (or memorize) that the maximum or minimum profit occurs on the edges–usually at a corner for the rigged setups of most linear programming problems in textbooks. Thinking about this led to several lovely technology enhancements. INSIGHT 1: Vary a point. During our first department meeting, I was suddenly dissatisfied with how I’d always introduced this idea to my classes. That unease and our play with the Desmos’ simplicity of adding sliders led me to try graphing a random ordered pair. I typed (a,b) on an input line, and Desmos asked if I wanted sliders for both variables. Sure, I thought (image below, file here). — See my ASIDE note below for a philosophical point on the creation of (a,b). — GeoGebra and the Nspire require one additional step to create/insert sliders, but GeoGebra’s naming conventions led to a smoother presentation–see below. BIG ADVANTAGE: While the Illuminations problem we were using had convenient vertices, we realized that students could now drag (a,b) anywhere on the graph (especially along the boundaries and to vertices of the feasible region) to determine coordinates. Establishing exact coordinates of those points still required plugging into equations and possibly solving systems of equations (a possible entry for CAS!). However discovered, critical coordinates were suddenly much easier to identify in any linear programming question. HUGE ADVANTAGE: Now that the point was variably defined, the Optimization Equation could be, too! Rewriting and entering the Optimation Equation as an expression in terms of a and b, I took advantage of Desmos being a calculator, not just a grapher. Notice the profit value on the left of the image. With this, users can drag (a,b) and see not only the coordinates of the point, but also the value of the profit at the point’s current location! Check out the live version here to see how easily Desmos updates this value as you drag the point. From this dynamic setup, I believe students now can learn several powerful ideas through experimentation that traditionally would have been told/memorized. STUDENT DISCOVERIES: 1. Drag (a,b) anywhere in the feasible region. Not surprisingly, the profit’s value varies with (a,b)‘s location. 2. The profit appears to be be constant along the edges. Confirm this by dragging (a,b) steadily along any edge of the feasible region. 3. While there are many values the profit could assume in the feasible region, some quick experimentation suggests that the largest and smallest profit values occur at the vertices of the feasible region. 4. DEEPER: While point 3 is true, many teachers and textbooks mistakenly proclaim that solutions occur only at vertices. In fact, it is technically possible for a problem to have an infinite number optimal solutions. This realization is discussed further in the CONCLUSION. ASIDE: I was initially surprised that the variable point on the Desmos graph was directly draggable. From a purist’s perspective, this troubled me because the location of the point depends on the values of the sliders. That said, I shouldn’t be able to move the point and change the values of its defining sliders. Still, the simplicity of what I was able to do with the problem as a result of this quickly led me to forgive the two-way dependency relationships between Desmos’ sliders and the objects they define. GEOGEBRA’S VERSION: In some ways, this result was even easier to create on GeoGebra. After graphing the feasible region, I selected the Point tool and clicked once on the graph. Voila! The variable point was fully defined. This avoids the purist issue I raised in the ASIDE above. As a bonus, the point was also named. Unlike Desmos, GeoGebra permits multi-character function names. Defining $Profit(x,y)=15x+30y$ and entering $Profit(A)$ allowed me to see the profit value change as I dragged point A as I did in the Desmos solution. The $Profit(A)$ value was dynamically computed in GeoGebra as a number value in its Algebra screen. A live version of this construction is on GeoGebraTube here. At first, I wasn’t sure if the last command–entering a single term into a multivariable term–would work, but since A was a multivariable point, GeoGebra nicely handled the transition. Dragging A around the feasible region updated the current profit value just as easily as Desmos did. INSIGHT 2: Slide a line. OK, this last point is really an adaptation of a technique I learned from some of my mentors when I started teaching years ago, but how I will use it in the future is much cleaner and more expedient. I thought line slides were a commonly known technique for solving linear programming problems, but conversations with some of my colleagues have convinced me that not everyone knows the approach. Recall that each point in the feasible region has its own profit value. Instead of sliding a point to determine a profit, why not pick a particular profit and determine all points with that profit? As an example, if you wanted to see all points that had a profit of$100, the Optimization Equation becomes $Profit=100=15x+30y$.  A graph of this line (in solid purple below) passes through the feasible region.  All points on this line within the feasible region are the values where Apu could build dirt bikes and get a profit of $100. (Of course, only integer ordered pairs are realistic.) You could replace the 100 in the equation with different values and repeat the investigation. But if you’re thinking already about the dynamic power of the software, I hope you will have realized that you could define profit as a slider to scan through lots of different solutions with ease after you reset the slider’s bounds. One instance is shown below; a live Desmos version is here. Geogebra and the Nspire set up the same way except you must define their slider before you define the line. Both allow you to define the slider as “profit” instead of just “p”. CONCLUSIONS: From here, hopefully it is easy to extend Student Discovery 3 from above. By changing the P slider, you see a series of parallel lines (prove this!). As the value of P grows, the line goes up in this Illuminations problem. Through a little experimentation, it should be obvious that as P rises , the last time the profit line touches the feasible region will be at a vertex. Experiment with the P slider here to convince yourself that the maximum profit for this problem is$165 at the point $(x,y)=(3,4)$.  Apu should make 3 Riders and 4 Rovers to maximize profit.  Similarly (and obviously), Apu’s minimum profit is $0 at $(x,y)=(0,0)$ by making no dirt bikes. While not applicable in this particular problem, I hope you can see that if an edge of the feasible region for some linear programming problem was parallel to the line defined by the corresponding Optimization Equation, then all points along that edge potentially would be optimal solutions with the same Optimization Equation output. This is the point I was trying to make in Student Discovery 4. In the end, Desmos, GeoGebra, and the TI-Nspire all have the ability to create dynamic learning environments in which students can explore linear programming situations and their optimization solutions, albeit with slightly different syntax. In the end, I believe these any of these approaches can make learning linear programming much more experimental and meaningful. ## Old school integral This isn’t going to be one of my typical posts, but I just cracked a challenging indefinite integral and wanted to share. I made a mistake solving a calculus problem a few weeks ago and ended up at an integral that looked pretty simple. I tried several approaches and found many dead ends before finally getting a breakthrough. Rather than just giving a proof, I thought I’d share my thought process in hopes that some students just learning integration techniques might see some different ways to attack a problem and learn to persevere through difficult times. In my opinion, most students taking a calculus class would never encounter this problem. The work that follows is clear evidence why everyone doing math should have access to CAS (or tables of integrals when CAS aren’t available). Here’s the problem: Integrate $\int \left( x^2 \cdot \sqrt{1+x^2} \right) dx$. For convenience, I’m going to ignore in this post the random constant that appears with indefinite integrals. While there’s no single algebraic technique that will work for all integrals, sometimes there are clues to suggest productive approaches. In this case, the square root of a binomial involving a constant and a squared variable term suggests a trig substitution. From trig identities, I knew $tan^2 \theta + 1 = sec^2 \theta$, so my first attempt was to let $x=tan \theta$, which gives $dx=sec^2 \theta d\theta$. Substituting these leads to $(tan \theta)'=sec^2 \theta$, claiming two secants for the differential in a reversed chain rule, but left a single secant in the expression, so I couldn’t make the trig identities work because odd numbers of trigs don’t convert easily using Pythagorean identities. Then I tried using $(sec \theta)'=sec \theta \cdot tan \theta$, leaving a single tangent after accounting for the potential differential–the same problem as before. A straightforward trig identity wasn’t going to do the trick. Then I recognized that the derivative of the root’s interior is $2x$. It was not the exterior $x^2$, but perhaps integration by parts would work. I tried $u=x \longrightarrow u'=dx$ and $v'=x\sqrt{1+x^2} dx \longrightarrow v=\frac{1}{2} \left( 1+x^2 \right)^{3/2} \cdot \frac{2}{3}$. Rewriting the original integral gave The remaining integral still suggested a trig substitution, so I again tried $x =tan \theta$ to get but the odd number of secants led me to the same dead end from trigonometric identities that stopped my original attempt. I tried a few other variations on these themes, but nothing seemed to work. That’s when I wondered if the integral even had a closed form solution. Lots of simple looking integrals don’t work out nicely; perhaps this was one of them. Plugging the integral into my Nspire CAS gave the following. OK, now I was frustrated. The solution wasn’t particularly pretty, but a closed form definitely existed. The logarithm was curious, but I was heartened by the middle term I had seen with a different coefficient in my integration by parts approach. I had other things to do, so I employed another good problem solving strategy: I quit working on it for a while. Sometimes you need to allow your sub-conscious to chew on an idea for a spell. I made a note about the integral on my To Do list and walked away. As often happens to me on more challenging problems, I woke this morning with a new idea. I was still convinced that trig substitutions should work in some way, but my years of teaching AP Calculus and its curricular restrictions had blinded me to other possibilities. Why not try a hyperbolic trig substitution? In many ways, hyperbolic trig is easier to manipulate than circular trig. I knew $\frac{d}{dt}cosh(t)=sinh(t)$ and $\frac{d}{dt}sinh(t)=cosh(t)$, and the hyperbolic identity $cosh^2t - sinh^2t=1 \longrightarrow cosh^2t=1+sinh^2t$. (In case you haven’t worked with hyperbolic trig functions before, you can prove these for yourself using the definitions of hyperbolic sine and cosine: $cosh(x)=\frac{1}{2}\left( e^x + e^{-x} \right)$ and $sinh(x)=\frac{1}{2}\left( e^x - e^{-x} \right)$.) So, $x=sinh(A) \longrightarrow dx=cosh(A) dA$, and substitution gives Jackpot! I was down to an even number of (hyperbolic) trig functions, so Pythagorean identities should help me revise my latest expression into some workable form. To accomplish this, I employed a few more hyperbolic trig identities: 1. $sinh(2A)=2sinh(A)cosh(A)$ 2. $cosh(2A)=cosh^2(A)+sinh^2(A)$ 3. $cosh^2(A) = \frac{1}{2}(cosh(2A)+1)$ 4. $sinh^2(A) = \frac{1}{2}(cosh(2A)-1)$ (All of these can be proven using the definitions of sinh and cosh above. I encourage you to do so if you haven’t worked much with hyperbolic trig before. I’ve always liked the close parallels between the forms of circular and hyperbolic trig relationships and identities.) If you want to evaluate $\int x^2 \sqrt{x^2+1} dx$ yourself, do so before reading any further. Using equations 3 & 4, expanding, and then equation 3 again turns the integral into something that can be integrated directly. The integral was finally solved! I then used equations 1 & 2 to rewrite the expression back into hyperbolic functions of A only. The integral was solved using the substitution $x=sinhA \longrightarrow A=sinh^{-1}x$ and (using $cosh^2A-sinh^2A=1$), $coshA=\sqrt{x^2+1}$. Substituting back gave: but that didn’t match what my CAS had given. I could have walked away, but I had to know if I had made an error someplace or just had found a different expression for the same quantity. I knew the inverse sinh could be replaced with a logarithm via a quadratic expression in $e^x$. Well, that explained the presence of the logarithm in the CAS solution, but I was still worried by the cubic in my second term and the fact that my first two terms were a sum whereas the CAS’s solution’s comparable terms were a difference. But as a former student once said, “If you take care of the math, the math will take care of you.” These expressions had to be the same, so I needed to complete one more identity–algebraic this time. Factoring, rewriting, and re-expanding did the trick. What a fun problem (for me) this turned out to be. It’s absolutely not worth the effort to do this every time when a CAS or integral table can drop the solution so much more quickly, but it’s also deeply satisfying to me to know why the form of the solution is what it is. It’s also nice to know that I found not one, but three different forms of the solution. Morals: Never give up. Trust your instincts. Never give up. Try lots of variations on your instincts. And never give up! ## Fun with Series Two days ago, one of my students (P) wandered into my room after school to share a problem he had encountered at the 2013 Walton MathFest, but didn’t know how to crack. We found one solution. I’d love to hear if anyone discovers a different approach. Here’s our answer. PROBLEM: What is the sum of $\displaystyle \sum_{n=1}^{\infty} \left( \frac{n^2}{2^n} \right) = \frac{1^2}{2^1} + \frac{2^2}{2^2} + \frac{3^2}{3^3} + ...$ ? Without the $n^2$, this would be a simple geometric series, but the quadratic and exponential terms can’t be combined in any way we knew, so the solution must require rewriting. After some thought, we remembered that perfect squares can be found by adding odd integers. I suggested rewriting the series as where each column adds to the one of the terms in the original series. Each row was now a geometric series which we knew how to sum. That meant we could rewrite the original series as We had lost the quadratic term, but we still couldn’t sum the series with both a linear and an exponential term. At this point, P asked if we could use the same approach to rewrite the series again. Because the numerators were all odd numbers and each could be written as a sum of 1 and some number of 2s, we got where each column now added to the one of the terms in our secondary series. Each row was again a geometric series, allowing us to rewrite the secondary series as Ignoring the first term, this was finally a single geometric series, and we had found the sum. Does anyone have another way? That was fun. Thanks, P. ## Still imbalanced ‘Blogging and Twitter are all about sharing ideas, and hopefully growing as a result. In this, my 100th post, I share a creation I made in response to Paul Salomon‘s Imbalance Challenge. As he notes on his ‘blog, these challenges emerged when he wondered how he could shift his young students to some deeper thinking from their conversations about equivalence. From the image below, can you rank circle, square, and triangle from lightest to heaviest? Thanks to those who share with me. I hope some of my offerings help others think and grow, too. ## Algebra Explorations Before Algebra Here’s a short post to share two great tools for students to learn algebra–long before any formal algebra course and without seeming like you’re even learning algebra. ACTIVITY ONE: The first is a phenomenal recent set of posts on Imbalance Problems (here and here) from Paul Salomon. If you’re on Twitter, interested in math or math education, and haven’t already, you should definitely follow him (@lostinrecursion). Paul is (or was) hosting an Imbalance Problem competition (mentioned at the end of his first post on imbalance problems), but the following image from his post gives the general idea. Can you figure these out? Better yet, can you write some of your own? Can you encourage your students (or kids) to create some? The process of thinking about the values of the unknown measures of the circles, squares, and triangles necessary to create these puzzles lies at the very heart of the concept of unknown variables that is so critical to algebraic reasoning. Best of all, this feels like a game, and puzzle solvers don’t even realize they’re learning algebra. ACTIVITY TWO: Several months ago, a Westminster alumnus (Thanks, Phillip!) suggested an iPad app that my 3rd grade daughter instantly fell in love with–Dragon Box. The app is available for$5.99 on both iOS and Android platforms. Jonathan H. Liu (@jonathanhliu) wrote such a great overview of the play on DragonBox for Wired.  Rather than trying to imitate his post, I recommend you read his review:  DragonBox: Algebra Beats Angry Birds.

The play of the game is great for teaching algebraic skills, again without ever seeming like it’s teaching as much as it is.  As a parent, I was THRILLED to see an educational game pull my daughter in so effectively and completely.  A high point came when I was driving carpool last week and my daughter recommended to a friend who was wondering what app would be fun to play for the ride, “Try DragonBox.  It’s fun!”

As a math teacher, I certainly can appreciate the game’s end-of-level challenges to get the box alone (solve for a variable) in the right number of moves (efficiency) using the right number of cards (also efficiency).  Still, there were a few times when I noted that a particular scenario could have been solved using a different sequence of equally-efficient moves that were not appropriately acknowledged as such by the software.  My teacher side wasn’t particularly pleased with the only-one-way-earns-top-recognition approach of the app, especially when other alternatives are equally valid.  Too many times, I fear students are faced with similar scenarios in their math classes.  Efficiency and elegance are certainly valuable skills in mathematics, but I think we too often try to impose rigor on young learners long before they have achieved basic understanding.

My grousing aside, my daughter and her friends weren’t bothered at all by the rare times where I had identified alternatives–I don’t think they even noticed.  Back to my point about too much emphasis on “the ‘right’ way” too early, I decided not to mention it to them.  As Mark Twain noted, I decided not “to let schooling get in the way of [their] education.”

CONCLUSION:  I hope you get a kick out of Paul’s imbalance problems (no matter what your age) and DragonBox if you have some younger kids around.  As always, make learning fun and not obvious–your charges will learn in spite of themselves!

## Exploring Sequences and Lines

Here’s another favorite problem that could be used for any middle or high school students who’ve been exposed to both arithmetic sequences and linear equations.

There is a family of lines, $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence.  What do all members of this family have in common?

As with most great math problems, the problem is simply stated and can be approached from many different perspectives.  In the end, this one isn’t that difficult to crack, but the solution (at least to me) is not intuitively obvious from the problem statement. If you’ve not seen this before, please try it before reading further.

WARNING:  Problem Solution Follows

MOST COMMON STUDENT APPROACH:  Given the problem’s phrasing, most begin by writing out a few possible equations using different arithmetic sequences, typically with positive common differences.  After a few tries, most get a little frustrated as no obvious pattern emerges from the equations.

It is this point that is typically the most difficult for me as a teacher.  I want to help anyone who struggles, but “help” often means allowing others the room to struggle and to find ways of solving problems on their own.  Only when a student or group gets really frustrated do I sometimes ask, “Are there any other ways you can look at this problem or at your work?”

Eventually, most decide to graph their equations to see if anything pops out.  The following graph was submitted this past semester by one of my precalculus students using the free online Desmos calculator.

Two lines intersecting in a point is common.  Three or more in a single point almost always indicates something interesting.  Seven lines through a single point is screaming for attention!  From this graph, all lines in this family apparently contain the point (1,-2).  That seems a bit random until you investigate further, but pattern recognition is just half of the battle for a mathematician.  Now that something interesting has been discovered, a mathematician wants to know if this is a fluke or something inherent to all members of the family.

NOTE:  All graphs my students have produced over the years have always contained specific equations.  I don’t know that it’s any more enlightening, but I did create on Desmos a slider version of the graphs of this family with one slider for the initial term (A) and a second for its common difference (d).

UNIQUE SOLUTION METHODS FROM STUDENTS:

All successful solutions eventually rewrite the sequence $\left\{ A,B,C \right\}$ to $\left\{ A,A+d,A+2d \right\}$ where d is the common difference for a generic arithmetic sequence with initial term A.

Method I: After discovering the common point of intersection, most plug it into the left side of the equation and simplify to get

$Ax+By+C=A\cdot 1+\left( A+d\right)\cdot -2+\left( A+2d \right)=0$.

Because the left side reduces to zero for all generic arithmetic sequences, $\left\{ A,A+d,A+2d \right\}$, (1,-2) must be common to all members of this family.

A few students aren’t comfortable dealing with 0=0, so these tend to plug in $x=1$ and solve for y to get $y=-2$, proving that the y-coordinate for $x=1$ for all members of this family is always -2.

Method II:  A few students every year work algebraically from properties of arithmetic sequences.  For any arithmetic sequence, $\left\{ A,B,C \right\}$, $\frac{A+C}{2}=B$.  This rewrites to $1\cdot A-2\cdot B+C=0$, so whenever $\left( x,y \right)=\left(1,-2 \right)$, then $Ax+By+C=0$ is a fundamental property of all arithmetic sequences.

Personally, I think this method gets closest to explaining why the point (1,-2) is the common characteristic of this family.

Method III:  This year, I had a student take an approach I’d never seen before.  She defined one arithmetic sequence as $\left\{ a,a+d,a+2d \right\}$ and another as $\left\{ m,m+n,m+2n \right\}$ for any real values of a,d,m, and n.  This leads to a system of equations: $a\cdot x+(a+d)\cdot y+(a+2d)=0$ and $m\cdot x+(m+n)\cdot y+(m+2n)=0$ .  If you have some younger students or if all the variables make you nervous, the solution is available via Wolfram Alpha.

Still, this system is pretty easy to manipulate.  If you multiply the first equation by m and the second by a, the x-terms will eliminate with subtraction, giving

$m\cdot((a+d)\cdot y+(a+2d))-a\cdot((m+n)\cdot y+(m+2n))=0$.

Solving for y shows that all of the coefficients simplify surprisingly easily.

$((ma+md)-(am+an))\cdot y=-(ma+2md)+(am+2an)$
$(md-an)\cdot y = -2\cdot (md-an)\Longrightarrow y=-2$

From here, determining $x=1$ is easy, proving the relationship.

SOLUTIONS and APPROACHES NOT YET OFFERED BY STUDENTS:

Approach A:  High school students don’t often think about ways to simplify problem situations, especially at the beginning of problems.  One way I did that for this problem in later class discussions was to recognize that it one of the terms in the arithmetic sequence was 0, you didn’t need to deal with nearly as many terms. For example, if your sequence was ${1,0,-1}$, the linear equation would be $x-1=0$.  Similarly, the sequence $\left\{ 01,2 \right\}$ leads to $y+2=0$.  Obviously, the only thing these two lines have in common is the point (1,-2).  A proof of the property must still be established, but this is one of the fastest ways I’ve seen to identify the central property.

Approach B:  A purely algebraic approach to this problem could redefine the arithmetic sequence as $\left\{ a,a+d,a+2d\right\}$as before, giving:

$a\cdot x+(a+d)\cdot y+(a+2d)=0$

Collecting like terms gives

$(x+y+1)\cdot a+(y+2)\cdot d=0$.

The values of a and d must remain as parameters to include all possible arithmetic sequences.  Because the equation always equals 0, the coefficients of a and d are both 0, making $y=-2$ (for the coefficient of d) and therefore $x=1$.

EXTENSION:

We once had a test question at the end of the unit containing this exercise.  Basically, it reminded students that they had discovered that all lines $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence contained the point (1,-2).  It then asked for an equation of a family of linear functions using the same arithmetic $\left\{ A, B, C \right\}$ that all contained the point (1,2).

The two most common responses we’ve seen involve a reflection or a vertical translation.  (1,-2) can become (1,2) by reflecting over the x-axis, so making the y-values negative would do the trick:  $Ax-By+C=0$.  Similarly, (1,-2) can become (1,2) by translating up 4 units, giving $Ax+B(y-4)+C=0$.