1,-3,9,-27,... Investigation

No matter where you introduce a lesson on geometric sequences, we can always begin with definitions, rules and formulas. 

The alternative is to build on student intuition and natural curiosity by asking them to write their own observations and questions they would like to have answered.

Imaginary Scenario (or is it?)

Jack: Mom, all the terms are just powers of 3 or their opposites, right?

Mom (Jane): Write your hypothesis, test it and let me know.

If your students or your son is not 15 year old Jack Andraka, here are some suggestions...


1.  What are the next 3 terms?
2.  If the 99th term is x, write an expression for the 100th term? (Recursive thinking)
3.  Which terms are positive? Negative?
4.  Write an expression for the nth term.
5.  How would we graph the sequence?
6.  Are the terms of the sequence increasing? Decreasing? Both? Neither?
7.  Which terms of the sequence are greater than a million? A trillion? Less than -1000000?

Another Imaginary Scenario (or is it?)

Uh, show me where this topic is in the CCSSM.
Uh, where does it say I have to ask all these questions?
Jack who?

Sent from my Verizon Wireless 4GLTE Phone

Classic Exponent Challenge for SATs, Algebra 2, Math Contests...

Don't forget to register for the upcoming 2nd MathNotations Free Online Contest for secondary students. Click here for more info.



The first 4 terms of a sequence are 2, 6, 18, and 54.
Each term after that is three times the preceding term.
If the sum of the 49th, 50th and 51st terms of this sequence is expressed as k⋅3⁴⁹, then k = ?

Click Read more to see the answer, solution, discussion...


Answer: 26/3

Suggested Solution
The first three terms can be written as
2(3⁰), 2(3¹) and 2(3²). (***)
In general, the nth term is 2(3^n-1).
The sum of the 3 desired terms would then be 2(3⁴⁸) + 2(3⁴⁹) +2(3⁵⁰). Factoring out 3⁴⁹, we obtain 3⁴⁹(2/3 + 2 + 6) = (26/3)(3⁴⁹), so k = 26/3.

Comments
(1) Too hard for SATs? Similar (but slightly easier) problems have appeared on the test.
(2) Could students use the "Make it simpler strategy" here to reduce the problem to the sum of just the first three terms? But this is the essence of geometric sequences (or exponential functions):
From (***) above, this sum would be
2(3⁰) + 2(3¹) + 2(3²) = 2 + 6 + 18 = 3(2/3 + 2 +6) = 3(26/3). The coefficient 26/3 would be the same for any three consecutive terms! Is this concept/technique worth developing?

...Read more

Squeezing Circles Into the Corner: An Infinite Sequence Investigation in Geometry

Another summer diversion from geometry...

The number of variations for tangent circles is endless and this is one of my all-time favorites. Math contests and SATs seem to have a preference for circles inscribed in squares or tangent circle problems and this one is along those lines. However, the real payoff comes from developing recursive thinking leading to an infinite geometric sequence and its sum! Students will be asked to intuitively "guess" the value of this infinite sum and to then verify their conjecture. Proving it requires nothing more than the classic formula for the sum of an infinite geometric series but, at the outset, this problem is eminently suitable for your geometry classes. Don't hesitate to use it in your "regular" classes. Questions that are deemed appropriate only for honors classes are often suitable for most students if the groundwork is laid (background, examples, etc.) and hints are given strategically.

PART I In the diagram above the larger circle has radius 1, the two circles are tangent to each other and to the two perpendicular segments (you can think of the larger circle being inscribed in a square if you wish).

(a) Make a conjecture from the diagram without computing: The ratio of the radius of the smaller circle to the larger is approximately
(A) 0.05 (B) 0.15 (C) 0.25 (D) 0.35 (E) 0.5
Note: This part may be omitted.

(b) Show that the radius of the smaller circle is exactly (√2 - 1)² = 3 - 2√2
How was your conjecture?
Note: Your decision about giving them the result like this. Obviously if they see part (b) on a worksheet, their estimate in part (a) will be pretty good! My intent was to focus on the method. Of course, feel free to rephrase this.

PART II
Of course we will not stop at 2 circles! Squeeze a third circle into the corner between the 2nd circle and the right angle. Determine its radius by using the result from part (a). [The key here is to think ratios!]

PART III
If we label the radius of the largest circle R₁, the radius of the 2nd circle R₂, the radius of the 3rd circle R₃, etc., we can now define an infinite sequence of these radii.
(a) Find a formula for the nth term of this sequence, n = 1,2,3,..

(b) What is the mathematical terminology for this type of sequence?

(c) Think intuitively here: From the diagram, what should be the "sum" of the original radius R₁ = 1 and the diameters of the remaining infinite collection of circles. [Another formulation: As n-->∞, this sum approaches what number?]

(d) Using the formula for the sum of an infinite geometric series, verify your conjecture in (c).

Comments:

• As always, feel free to use this with your students and revise as you see fit. However, pls use the attribution in the Creative Commons License as indicated in the sidebar.
• Finding the radius of the 2nd circle is a challenge by itself and the problem could stop there. The extensions can be assigned as a long-term project or for those wishing to do extra credit. I always liked having additional challenges for the students who were capable of going further, although relating this problem to geometric sequences or series is of importance. Of course, I am well aware of time constraints faced by the instructor.
• Your thoughts...
  

Mortgages - Third installment

[Update as of 6-17-07: At the bottom you will now see 3 screenshots from the TI-84 showing all of the formulas used for this series of mortgage activities and the input screen for the built-in Finance Application on the TI-84 that can be used to determine the monthly mortgage payment. The first 2 screens overlap, i.e., the 2nd screen contains part of the first screen and the 4th function, Y4. You will need to refer to the index of variables below to make sense of all this. There are more details below.]

The following is the 3rd and possibly the last in this particular series of classroom activities. All three should be assigned for complete effect:
Part I: Taking the Magic Out of Mortgages
Part II: Puff the Magic Mortgage

Thought I forgot to finish this activity?
Well, with the school year over for some and ending for others, here's Mortgages Part III to think about as we look forward to making our monthly payments during the summer and plan enrichment classroom activities for the fall and spring. Part III is more ambitious and requires more sophistication on the part of the Algebra 2, Advanced Algebra or Precalculus student. As always, I am attempting to provide a completely developed enrichment lesson ready to use or modify as needed. You may want to bookmark this and return to it when teaching this unit next year.
The goals here are:
(a) Providing a more challenging application of exponential functions and their relation to geometric sequences and series
(b) Systematic development of the formulas for the equalized monthly mortgage payment as well as the portion of the monthly payment that goes toward paying off the principal, etc.

This is an activity that is particularly suited for block scheduling. If begun in a 40-45 minute period, the lesson will probably run over two periods or the last few parts can be assigned for homework. Another effective approach is to give this as a long-term individual or group project. In this case, I would recommend combining all three Mortgage activities.

STUDENT ACTIVITY
In the previous activity, you should have observed that the sequence of data values in the Y₁ column formed a geometric sequence with common ratio 1+I, where I was the interest rate per payment period (decimal form). It's time to derive this mathematically and see how the other columns were generated and how some of those famous mortgage formulas came to be. Did you figure out that Y₁ contained the amounts labeled P_x below?

The following is an index of the variables we will use . I'm using uppercase variables and X for ease of entry when instructed to enter these formulas into your graphing calculator. Note that the discussion below answers the questions from the previous activity regarding the meanings of the Y-columns in the calculator.

P = Original amount of Loan (remember, it was $100 in the previous activity)
I = Rate of interest per payment (expressed as a decimal)
Note: E.g., if the bank is charging 6% annual rate on your loan, I = 6/12% or 1/2% = 0.005 per month!
Z = 1 + I (to make formulas easier to write and enter into the calculator, since 1+I appears frequently when doing compound interest)
N = number of payments (e.g., N = 360 for 12 payments a year over 30 years)
X = the index used for the xth payment
P_x = Amount of the xth monthly payment that goes toward reducing the principal
I_x = Monthly interest payment
A = Level (equal) monthly payment
U_x = Amount of debt (Unpaid amount) remaining after Xth payment


(1) Explain the meaning of the equation: P₁ + PI = P₂ + (P-P₁)I.
(2) Show that P₂ = P₁(1+I) by solving the equation in (1) for P₂.
(3) Explain why P₁ + PI = P₃ + (P - P₁ - P₂)I
(4) Show that P₃ = P₁(1+I)² by solving the equation in (3) for P₃ (after substituting for P₂ from (2)).

The results in questions (2) and (4) suggest the following general formula which can be verified by mathematical induction:
(**) P_x = P₁(1+I)^X-1.
Recall that P_x denotes the amount of the Xth payment that goes toward paying off the original loan amount P.

The next few parts require that you recall the formula for the sum of the first N terms of a geometric sequence. If you have forgotten it, research it or your instructor will review it.

(**) shows that the sequence P_x is a geometric sequence with first term P₁ and common ratio, 1+I (or Z).
(5) Explain why P = P₁ + P₂ + P₃ + ... + P_N
(6) Using (5) and the formula for the sum of the first N terms of a geometric sequence, show that P₁ = PI/((1+I)^N-1) = PI/(Z^N-1) where Z = 1+I.
(7) Use (6) to explain why A = PI/((1+I)^N-1) + PI.
(8) Simplify the result of (7) to derive:
A = PI(1+I)^N/((1+I)^N-1) = PIZ^N/(Z^N-1)
[Again, Z = 1+I]

(9) STORE the following values from the Home screen:
100 STO P
.1/12 STO I [10% annual rate divided by the number of payments during the year]
1+I STO Z
12 STO N

Note: If you haven't used the ALPHA key before, you will now! Remember: The variables listed above will store these constant values until you or some program changes them. Clearing the screen has no effect on stored variables.

(10) Enter the last formula for A (Z-form) from (8) into Y₁ in your graphing calculator. You may have to modify it slightly for entry purposes. The * symbol for multiplication is not necessary for most graphing calculators. Try it!

(11) Start a TABLE from X = 1 and display your TABLE. If entered correctly, the values for
X = 1 through 12 should all be the same. Why? Which column was this in Part II of the Mortgage Activity?

(12) Using ** and the formula for P₁ from (6) (the one in Z-form), write a formula for P_x in terms of P, I, Z, X and N. Enter this into Y₂. Display the TABLE starting from X = 1. Which column was this in Part II of the Mortgage Activity?

(13) Derive a formula for I_x using preceding results. Again, express it in terms of P, I, Z, X and N and enter this into Y₃. Which column was this in Part II of the Mortgage Activity? Explain why these values are decreasing.

(14) Derive a formula for U_x using preceding results. Again, express it in terms of P, I, Z, X and N and enter this into Y₄. Which column was this in Part II of the Mortgage Activity? Explain why these values are decreasing.

NEW!!
Below you will find 3 screenshots from the TI-84. The first 2 show the actual functions used to compute the 4 key quantities used for mortgage repayments. The 3rd screenshot shows the finance application screen (APPS, Finance, TVM Solver) used to input the actual data values used in this activity. Students will need to refer to the index of variables above to make sense of these functions. PMT (the monthly mortgage payment) was obtained by pressing ALPHA ENTER (SOLVE). One of the main goals of this series of activities was to show students how they could obtain the formulas that are hidden behind this 'cool' application. Ask your students to explain why PMT is displayed as a negative amount!

Y1 = The payment toward principal function, i.e., the portion of the xth monthly payment that is applied to the loan principal (increasing function)
Y2 = The monthly interest payment (decreasing function)
Y3 = The fixed monthly mortgage payment (constant function, thus the variable x does not appear)
Y4 = The debt function, i.e., the amount still owed on the principal after the xth payment (decreasing function)

Taking the Magic out of Mortgages Part I: Exponential Functions and Geometric Sequences to the Rescue

[Note: For an exceptionally clear and definitive exposition of all things financial, the best resource I have found is MoneyChimp. There are interactive calculators to thoroughly understand the concepts in this post and much much more. More importantly, for math nerds like me, the formulas are explained and, in some cases, derived. The mathematics is accurate and the analysis is excellent. Enjoy it!]

In Algebra 2 and Precalculus (or whatever it may be entitled in your local schools), students often do compound interest problems. Typically, the author of the text and/or the instructor will derive the formula for what your original investment of P dollars will be worth in t years, if interest is compounded n times per year for t years at an annual rate given by r (a decimal for this discussion):
Compound Interest Formula: A(t) = P(1+r/n)^nt.

This is a nice practical application of exponential functions, exponential growth in particular. A similar, but more sophisticated, concept applies to annuities and amortization of a mortgage (paying off debt over time in n equal payments). In both an annuity and a mortgage, the original amount of money (whether it's the amount invested or the debt you owe) generally decreases over time. In an annuity, you receive a fixed amount at the end of each period, whereas, in a mortgage, you pay a fixed amount. In an annuity, your original investment is earning (accruing) interest (it may be possible to 'live off' the interest and not touch the principal), while you are receiving periodic equal payments that are deducted from your account. A central concept in both annuities and mortgages is that that interest is applied before receiving an annuity payment or before making a mortgage payment.

The following is the first part of an activity introducing students to the mysteries of mortgage calculations. The fact that the formulas for monthly payments or the decreasing amount of debt seem very intricate lead many to believe that this topic is too sophisticated for most secondary math students. Just give them the formulas, mention that it is related to exponential functions and let them plug it all into their graphing calculators. We know most adults, other than those in the business of lending, punch the numbers into the computer and read the results. Before calculators, bankers would look it up in those mortgage tables on some well-worn-out card. This activity may demystify a little bit of this. Students need good algebra skills, knowledge of exponential properties and functions in particular, a basic knowledge of compound interest and background in geometric sequences and series (later on). I am well-aware that sensitivity is needed here for students whose parents do not own a home, however, all students can benefit from these ideas since these principles apply to far more than a monthly mortgage payment.


STUDENT ACTIVITY
You can find many excellent web resources for mortgage calculations. You can also find the actual formulas for all of this either in your text or in other sources. Most of us would probably use the built-in applications typically found on a graphing calculator or more likely use those free mortgage calculators all over the web. In this activity you will take an active role in the process of borrowing and lending and see what lies behind those sophisticated formulas.

In actual practice, mortgages can range from less than a hundred thousand into the millions of dollars. Therefore, these loans are typically repaid over 5, 10, 15, 20, 25 or 30 years to make the monthly payments more manageable. In this activity you will be borrowing a small amount and considering an oversimplified form of repayment, leading up to more general considerations.

You will be borrowing $100 from a reputable lender, Stan, The Mortgage Man.
Stan is charging the going rate at the moment, which is 10% compounded annually.

(a) If you repay the loan in one year, explain why your single payment would be $110.
(b) If you agree to repay the loan at the end of two years, in one single payment, explain why that single payment would be $121?

Discussion: Parts (a) and (b) should remind you of the compound interest formula you've learned:
One year re-payment: 100(1+0.1)¹
Two year re-payment: 100(1+0.1)²

Surely, increasing the payment schedule to TWO payments over two years or one year, cannot be that much more difficult? Let's find out...

(c) This time you will make two equal payments over two years. Stan gives you the repayment schedule: Two equal payments of $60. He explains it as follows: The loan (your debt) of $100 is divided into two equal payments of $50 each. The interest charges are $10 (10% of the amount you borrowed) on each payment. Explain the mistake that Stan is making (or is he trying to take advantage of an unsuspecting borrower who didn't pay attention in algebra?). We're not asking you to correct Stan's error here - just explain why his calculation is either wrong or unfair.

(d) Now that you figured out that the two equal annual payments should not be $60, we will tell you what the actual payments would be according to mortgage formulas:
Each annual payment is $57.62 (rounded to the nearest penny).
Show why these two payments correctly repay the loan of $100 and the interest that is due on each payment. Show method clearly. Use calculator as needed.

(e) Do you think you could figure out an algebraic way to determine those equal payments of $57.62? You are about to...
Let A represent the equal annual payments you will make.
At the end of the first year, before you make your payment, you owe $100(1.1) = $110.00. Now the fun begins:
(i) Represent, in terms of A, the amount of debt (loan + interest) you will owe AFTER you make your first payment.
(ii) Represent, in terms of A, the interest you will be charged by the end of year 2 BEFORE you make your final payment.
(iii) Represent, in terms of A, your debt, AFTER making your final payment.
(iv) What should be the numerical value of your debt AFTER making your final payment? Now, write an equation and solve for A. You should come up with $57.62!

(f) Are you up to the challenge of solving for the general formula for A given an original loan of $P at an annual interest rate of i (expressed as a decimal)? Of course, you are! For now, you only need to do this for TWO payments, just as we did in (e). Of course, your formula for A should be in terms of P and i.

More to follow...

Bringing a 'Series' of Wonders to the Calculus Classroom

The following may drive away most casual readers but it does describe what I try to do every day. One of my goals in starting this blog was to enable a dialogue for effective instructional strategies. My focus has generally been on middle and secondary school curriculum up to Algebra 2, bordering on Precalculus. Today I am sharing a different experience. I hope some of you will appreciate it beyond its technical aspects. Similar developments can be found in some textbooks and experienced teachers already do most of this but as this scenario is fresh in my mind, I thought I'd re-play it for you...

Although most Advanced Placement Calculus (BC) teachers are completing or have already completed the unit on infinite series, I would like to offer a view that I hope brings a sense of 'shock and awe' to the student of the 21st century who rarely has the time to stop and appreciate the beauty of our subject. To those who have been teaching this for a while, you may not quite feel this. However, I still get goosebumps when I observe student reactions as this unfolds in front of their eyes...

Assume that students already have a basic understanding of infinite series, the infinite geometric series in particular.

Consider the following three infinite geometric series:

1+1/2+1/4+1/8+... = 1/(1-1/2) = 2
1+1/3+1/9+1/27+... = 1/(1-1/3) = 3/2
1-1/4+1/16-1/64+... = 1/(1-(-1/4)) = 4/5

Just a collection of simple geometric series, boys and girls?
Genius is looking at an ordinary collection of objects and seeing something different. Some mathematician or mathematicians (research this and report back with their bios!) may have considered a reverse view of these series. Instead of the goal being a formula for the sum of the series, perhaps the goal was to represent a function in a different way. Step back into history...

Consider the general formula for the sum of all these series: 1/(1-r) provided r is between -1 and 1. Replace r by x, the variable we usually use for functions, and we can write:
1 + x + x² + x³ +... = 1/(1-x) provided x is between -1 and 1.

The 'polynomial' of infinite degree on the left is known as a power series in x. As long as x is between -1 and 1 (the interval of convergence), this 'equation' makes sense and allows us to use algebraic and calculus operations to represent other related functions. Think of how one might have felt when 'discovering' this and I'm just speculating here. The rational function 1/(1-x) is being represented by some kind of polynomial that never ends. Even though x= 1 is not in the interval of convergence, substituting leads to 1/0 = 1 + 1 + 1 + 1 +.... Hmm....
Let's try substitution on this representation.
Replace x by -x²:
(You can show the domain is unchanged)
1/(1 - (-x²)) = 1 + (-x²) + (-x²)² + (-x²)³ + ... OR
1/(1 + x²) = 1 - x² + x⁴ - x⁶ +...

Ok, let's integrate both sides (assuming it's legal to do so):
tan^-1(x) = x - (1/3)x³ + (1/5)x⁵ - (1/7)x⁷ + ... + C
Replacing x by 0, we see that C = 0.
Now, you'll have to accept this for the moment (to be proved later), equality holds for x = 1, even though 1 was not in the original interval of convergence! It is not unusual when integrating a power series to see the domain include one or both endpoints even though the original function excluded them!

Thus, tan^-1(1) = 1 - 1/3 + 1/5 - 1/7 + ...
Anyone recognize the left-hand side?
The bell rings...

Variations on Basic Themes: Digit Problems in the Key of A or G Minor?

Update: Answers, solutions have now been posted in the comments.

I thought of these variations on the well-known combinatorial problems involving 3-digit numbers that pop up frequently as I was teaching arithmetic and geometric series yesterday.
These questions are appropriate for grades 6-12 provided students are given definitions and some practice with arithmetic and geometric sequences, topics that are well within the abilities of middle schoolers. A quick intro to these sequences is all that is really needed OR, as I did below, they can be defined in the problem itself. Thus, these questions provide both practice in arithmetic skills and in combinatorial thinking. Of course, all the experienced or budding programmers out there can write simple code to have their graphing calculators count these, but that should only complement and verify their results, not replace the reasoning needed to solve them, unless these are used for a computer science class (even then, programmers should independently verify their code by solving the problems!).
These are not highly challenging and therefore can be used as Problems of the Day, for extra credit, or enrichment. Our readers will hopefully suggest other extensions and further variations (some are suggested below).

1. The digits of 246 form an arithmetic sequence from left to right because 4-2= 6-4. How many positive 3-digit integers satisfy this condition?


2. The digits of 248 form a geometric sequence from left to right because 4/2 = 8/4. How many positive 3-digit integers satisfy this condition?

Now, how could we make these more challenging? 4-digit numbers or will that make one or both easier, i.e., fewer possibilities? What if the digits were allowed to form these sequences in any order? BTW, I apologize for the music pun in the title. I hope you will respond to that with a positive tone!