This unit looks at two general equations associated with a sequence or pattern: (i) recurrence relations as general ways to see what happens between consecutive terms of a sequence; and (ii) the general term of a sequence as a function of the number of that term. The work in this unit is valid for any arithmetic progression, that is, any sequence where the difference between any pair of consecutive terms is the same.
Recurrence relations: Implicitly, recurrence relations are probably the first thing that students use when they are dealing with patterns. It seems that it is easier to see the difference(s) between consecutive terms of a pattern than to see the general term. This is because the difference pattern is simpler than the original pattern. For instance, in patterns involving quadratic equations, the recurrence relation involves just a linear relation.
So what is a recurrence relation? It is just an equation that links a number of terms of a pattern. Here we will restrict it to mean the general relation between any pair of terms. For instance, in the sequence 4, 5, 6, 7, …, the difference between any two terms is 1. So if f(n) is the value of the n-th term of the pattern/sequence, then f(n) = f(n-1) + 1.
At this point it is worth noting that we will use f(n) for the general term of the sequence here. You may prefer to use Tn or some other notation.
Back to recurrence relations, any arithmetic progression where the difference between consecutive terms is d, has recurrence relation f(n) = f(n-1) + d.
Although we only worry here about recurrence relations that involve two terms, it is interesting to note that the Fibonacci sequence has recurrence relation f(n) = f(n-1) + f(n-2).)
Furthermore, with this type of recurrence relations, if we know the recurrence relation of a sequence and we know its first term, then we know the sequence completely. This is because we can start with the first term and then generate each successive term by applying the recurrence relation.
So if we are given f(n) = f(n-1) + 1 and f(1) = 2, we get 2, 3, 4, ..., and if we are given f(n) = f(n-1) + 1 and f(1) = 4, we get 4, 5, 6, …. So the initial term plays an important role in the sequence.
The problems with guessing: It is important to note that students often start off with a table of values for a ‘physical’ situation and then ‘guess’ the general term. This guess may not be correct. For instance we were working with students with a pattern that started out as 8, 48, …. One group guessed a pattern of 8 × 6n-1. While this fits the two values that they had found exactly, it is not the general term of the pattern they were investigating. Because they had failed to look for more terms, or to see the actual physical situation was changing, they went off in the wrong direction. The same thing can happen with the problem situations here. Therefore, it is important that recurrence relations are grounded in an actual physical problem, and not in a table of values that may be guessed from small values of the physical pattern.
Algebra versus geometry: We have taken two approaches to the general term here. This is done to present students with a variety of representations of the general term, and to enable students to see that there is more than one way to approach a problem.
Our algebraic approach works by using patterns, where the general term is a multiple of n, and inserting an extra column in our table between the number of the term and the value of that term. We then divide the number into the value to find a result.
The geometric approach relies less on guess work. Patterns such as the ones that we are working with here can be represented as rectangles – at least, they can when we put two of them together. This approach works for arithmetic progressions and related sequences.
The patterns of this unit: We look at five patterns here that we have called Barrels, Odd Bricks, Even Bricks, Dominoes and House of Cards (see Copymasters 1.1 to 1.5). We’ll discuss Barrels first and then see that the others are essentially the same.
The sequence that we get with Barrels is 1, 3, 6, 10, and so on. You might recognise these as being numbers in Pascal’s Triangle – take the diagonal that starts three ones down from the top. But because the rows of the n-th term have 1, 2, 3, 4, …, n barrels, the n-th term of the barrels sequence is actually the sum of the first n whole numbers. So to find the number of barrels in the n-th term we just have to add the first n whole numbers. This is the method we exploit in the geometric method of Sessions 3.
Odd Bricks goes 1, 4, 9, …. Clearly these are just the square numbers so it’s not too hard to find the n-th term. But how do we make up the n-th element? We simply add 1 to 3 to 5 to …. So the n-th term is just the sum of the first n odd numbers.
The n-th term of the Even Bricks’ sequence is the sum 2 + 4 + 6 + … + 2n; the n-th term of the Dominoes’ sequence is the sum 3 + 5 + 7 + … + (2n + 1); and the n-th term of the House of Cards’ sequence is the sum 2 + 5 + 8 + … + (3n – 1).
All of the n-th terms here are the sums of arithmetic sequences or arithmetic progressions. The elements of the sums increase by a constant, d, as n increases each time. We summarise each of these below.
pattern | First term | d | General term | Sum = n-th term of the pattern |
Barrels | 1 | 1 | n | n(n+1)/2 |
Odd Bricks | 1 | 2 | 2n - 1 | n2 |
Even Bricks | 2 | 2 | 2n | n(n + 1) |
Dominoes | 3 | 2 | 2n + 1 | n(n + 2) |
House of Cards | 2 | 3 | 3n - 1 | n(3n + 1)/2 |
So the n-th term of each of these patterns can be found if you know how to sum an arithmetic progression. And this sum is just the sum of the first and the last terms by the number of terms divided by 2. And it’s the fact that the difference between each term of the sum is the same that enables them to fit together so nicely for the geometric argument of session 3.
Incidentally, Odd Bricks, Even Bricks and Dominoes all have the same recurrence relation but have different initial terms. This underlines the fact that a sequence of numbers needs both a recurrence relation and an initial term to be specified before it is defined uniquely.
Where to next?: This whole area leads out in many directions. As we said earlier, it is easier for students to find recurrence relations for patterns than it is for them to find general terms. This is often the case for researchers too. But general terms are obviously more useful. Hence researchers have found ways to obtain general terms directly from recurrence relations. General terms for the recurrence relations of this unit can be found by solving appropriate quadratics. There are also things called generating functions that come in handy for a broader range of situations.
Anyone interested in these developments should read books on a topic called Combinatorics. It’s an essential part of a lot of what is known as Discrete Mathematics. The area has an increasing number of valuable applications to increasingly complex situations. Colleges in the USA generally have a course in these topics so it’s easy enough to find a book on the area or to find information on the internet.
The learning opportunities in this unit can be differentiated by providing or removing support to students, and by varying the task requirements. Ways to differentiate include:
The focus of this unit is not related to a specific, real-world context. However, as a class, you might consider contexts that are reflected in each of the patterns and sequences presented (e.g. the building of a wall for a school buildings, stacks of cans of fizzy drink at a school event). Furthermore, at the conclusion of this unit, you may wish to explore real world applications of recurrence relations, and the general term of a sequence as a function of the number of that term.
Te reo Māori kupu such as tauira (pattern), raupapa (sequence), whārite (equation), and tūtohi (table (of data), chart) could be introduced in this unit and used throughout other mathematical learning.
See the bottom of the unit for links to each Copymaster.
In this session we take a look at several pattern situations. The aim here is to see that recurrence relations can be used to describe patterns and that tables can be generated by using these recurrence relations. The material in this session may take up to two lessons. You may decide that there are more situations here than you want to use in this unit.
Note that we have used f(n) to denote the n-th term of a sequence.
Consider Pattern 1: Barrels. We show a picture of the situation below (also available in Copymaster 1.1)
One of the important aspects of a pattern is the way that subsequent terms are generated from previous terms. In the pattern above to go from f(1) to f(2) we add 2; to go from f(2) to f(3) we add 3; to go from f(3) to f(4) we add 4; and so on. So f(2) = f(1) + 2; f(3) = f(2) + 3; f(4) = f(3) + 4; and so on. We can see this by looking at the pictures to see how new terms are constructed.
This last equation tells us how to get any term from the previous one. By putting in various values for n we can find the other equations that we have just looked at. So this equation is valuable and is given a special name. It is called a recurrence relation because it tells us about the recurring relation between consecutive terms. One use of this recurrence relation is that it enables us to produce a table showing the number of barrels in any term. (Though this gets a little tedious when the term number is large – like 100, say. We will see how to deal with this later in this unit.)
In this session, the class will use their tables to produce values of f(n).
Explain to students:
As you know from the last session, the recurrence relations are useful as a tool to generate a table of values. However they are of limited value if you want to find, say, f(100), as they have to be used an inordinate number of times. This could easily lead to arithmetical errors. What would be nice would be a formula for f(n) so that we could just plug in, say 100, and get f(100) with a minimum of counting. We’ll tackle this problem here.What we are about to do is just a matter of guessing. We will try to guess the formula for f(n). But we’ll try to do it systematically.
Return to Pattern 1, the Barrels. Work with students to see how you can go about finding the pattern from the table of values that has been produced.
It will help us if we look at the table and try to work from there. Notice that I’ve added a middle column that I think will help us and I’ve put in tall the values up to n = 6.
n | f(n) | |
1 | 1 | |
2 | 3 | |
3 | 6 | |
4 | 10 | |
5 | 15 | |
6 | 21 | |
What I want to do is to see how to get from n to f(n). To do this I’m going to use the middle column to work out f(n)/n. So let’s do that now. Use questioning and discussion to complete the table below.
n | f(n)/n | f(n) |
1 | 1 | 1 |
2 | 3/2 | 3 |
3 | 2 | 6 |
4 | 10/4=5/2 | 10 |
5 | 3 | 15 |
6 | 21/6=7/2 | 21 |
Can you guess what the pattern of the middle column is? [(n+1)/2]
Is this an easier pattern to find than the pattern for f(n)?
So how do we get f(n) from n and this pattern of the middle column? [Multiply the pattern of the middle column by n to get f(n).]
What is f(n) then? [f(n) = n(n+1)/2.]
In this session, that may take more than one lesson, we look at a geometrical approach to finding formulae for the n-th terms of sequences. Naturally the method won’t work for all patterns but it will work for the ones that we have here and for arithmetic progressions generally.
Let’s run through this again with 7 rows of Barrels (see Copymaster 3.2).
Here is the first Barrels picture (top picture).
Let’s run through this again with n rows of Barrels (see Copymaster 3.3).
Dear families and whānau,
Recently we have been looking at two general equations associated with a sequence or pattern: (i) recurrence relations as general ways to see what happens between consecutive terms of a sequence; and (ii) the general term of a sequence as a function of the number of that term. We have explored these using algeraic and geometric methods. Ask your child to share their learning with you.
Printed from https://nzmaths.co.nz/resource/why-and-how-general-terms at 11:00pm on the 11th May 2024