Triangular Number Links
AO elaboration and other teaching resources
Use algebra to simplify expressions
Use geometry to assist their algebra.
This problem is the last of four problems relating to triangular numbers. The others are Triangular Numbers, Number, Level 3, Counting Pills, Number, Level 4, and Square and Triangular Numbers, Algebra, Levels 5. Over the range of these problems we develop the idea of the triangular numbers leading to an algebraic formula for the nth triangular number.
We are concerned here about patterns, how to continue them and how to find the general term of a pattern. Here we are able to get the nth term (the general term) of the triangular numbers as an expression in n in a number of ways.
This problem is really an exercise in using algebra in the context of a pattern that has been investigated quite a bit over the other three units. Along the way we link algebra and geometry, and recall the difference of two squares as a useful tool.
The Problem
Mary knew that the 5th triangular number is 15 because it needs 15 counters to make the triangle.
But she didn’t know which of these expressions is equal to the nth triangular number, T(n). Which are and which aren’t and why?
(1) T(n) = 1 + 2 + 3 + 4 + . . . + (n – 3) + (n – 2) + (n – 1) + n;
(2) T(n) = ½n(n + 1);
(3) T(n) = 1 + 3 + 5 + … + (2n – 5) + (2n – 3) + (2n – 1).
(4) T(n) = n^{2} – (n – 1)^{2} + (n – 2)^{2} – (n – 3)^{2} + . . . + 4^{2}  3^{2} + 2^{2} – 1^{2};
(5) T(n) = [(n + 1)^{2} + n^{2} + (n – 1)^{2} + ... + 1^{2}] – [(n^{2} + 2) + ((n – 1)^{2} + 2) + ((n – 2)^{2} + 2) + … + (1^{2} + 2) + 2];
(6) T(n) = T(n 1) + n and T(1) = 1.
Teaching sequence

Introduce the problem to the class.

Brainstorm ideas for approaching each part of the problem.

As the students work on the problem in pairs you might ask the following questions to extend their thinking:
What strategies might help you to find the answer?
How can you use your knowledge about algebra here?
Can you see any patterns that might help?
How does your geometric knowledge help? 
Share the students’ answers. Ask them to explain their reasoning. Talk about what is needed to write down a complete solution to the problem.

Ask students to write up their method of solution. Check that they have correctly used the main ideas of the argument.

Get as many of the students to work on the Extension problem as possible. Actually this could be done in conjunction with the original problem.

Discuss the extension problem with the whole class. Let them write it up another solution for each of the parts of the problem.

Keep a chart of the different ways of solving each part of the problem. Add to it over the next couple of weeks as ideas come to you and the class.
Extension to the problem
How many ways can you show that (1), (2) and (4) are triangular numbers?
Solution
Note that in a question like this you have to do one of two things. If you believe an expression is wrong, then it is often easiest to show that it is wrong for one particular value of n. On the other hand, if you believe that it is right, then you have to justify this by showing that it is equal to an expression that you know is right.

This is an expression for the nth triangular number so we have to prove that this is the case. We do this by constructing the nth triangular number in counters.
The number of counters in the bottom row of the nth triangular number is n. The number of counters in the row above that is n – 1. The number of counters in the row above that is n – 2. This continues until you get to the top row with 1 counter. So
T(n) = n + (n – 1) + (n – 2) + … + 2 + 1.

This is an expression for the nth triangular number so we have to prove that this is the case. We use an algebraic approach for this one.
Recall that the nth triangular number is formed by adding the first n numbers.
To show that (1) holds we use the backward/forward method from Square and Trianglular Numbers, Level 5. Suppose that the nth triangular number is T(n). Then
T(n) = 1 + 2 + 3 + 4 + . . . . + (n – 3) + (n – 2) + (n – 1) + n
T(n) = n + (n – 1) + (n  2) + (n – 3) + . . . . + 4 + 3 + 2 + 1
Then adding the two lines together gives
2 T(n) = (n+1) + (n+1) + (n+1) + (n+1) + . . . . + (n+1) + (n+1) + (n+1) + (n+1)
But there are n terms here so the right hand side of this expression is n x (n + 1). So
2 T(n) = n x (n + 1)
And this gives
T(n) = ½n(n + 1).

This is not an expression for the nth triangular number so we only have to find one value of n for which it is wrong.
Put n = 2. Then becomes 1 + 3 + 5 + … + (2n – 5) + (2n – 3) + (2n – 1) = 1 + 3 = 4. But the second triangular number is 1 + 2 = 3. So expression is not equal to T(n).

This is an expression for some nth triangular numbers so we have to prove that this is the case. We use an algebraic approach for this one.
To do this, we need to know the result for the difference of two squares.
The difference of 2 squares result is a^{2} – b^{2} = (a – b)(a + b). Geometrically it can be deduced as follows. Subtract a square of size b from a square of size a.
The difference is two rectangles. From the diagram we can see that dimensions (a – b) by b and the other (a – b) by a.
So the total area of the two rectangles is (a – b) x (a + b).
This show that a^{2} – b^{2} = (a – b)(a + b).
Now how are we going to use that on the expression for T(n)? Well, first note that the expression comes in pairs of squares one of which is subtracted form the other. For instance, the first pair is n^{2} – (n – 1)^{2}. Since this is the difference between two squares we can apply a^{2} – b^{2} = (a – b)(a + b). In the first pair a = n and b = n – 1. So
n^{2} – (n – 1)^{2} = [n – (n – 1)][n + (n – 1)] = n + (n – 1).
Since this is two of the terms to be found as the last number in the sum that makes up T(n) in (1), that’s a good start.
Similarly, using the difference of two squares we can show that (n – 2)^{2} – (n – 3)^{2} = 2n – 5 = (n – 2) + (n – 3). And that’s another two terms.
Each difference pair down to 2^{2} – 1^{2} = 2 + 1, contributes a pair of terms in T(n). So it’s looking good for the expression of (4) to be equal to T(n). But just look a bit carefully. If we start from the low end and work up we get
(2^{2} – 1^{2})^{ }+ (4^{2}  3^{2})^{ }+ (6^{2}  5^{2})^{ }+ (8^{2}  7^{2}) + …
= (1 + 2) + (3 + 4) + (5 + 6) + (7 + 8) + …
And here is the problem. We always get a pair of numbers from each bracket. So at the end of the day we don’t get
1 + 2 + 3 + 4 + . . . + (n – 3) + (n – 2) + (n – 1) + n
for every n. We only get it for n even. So the expression in (4) is only equal to T(n) if n is even.

This is not an expression for the nth triangular number so we only have to find one value of n for which it is wrong.
Try n = 1. Then [(n + 1)^{2} + n^{2} + (n – 1)^{2} + ... + 1^{2}] – [(n^{2} + 2) + ((n – 1)^{2} + 2) +
((n – 2)^{2} + 2) + … + (1^{2} + 2) + 2], becomes [2^{2} + 1^{2}] – [3 + 2] = 0.

This is an expression for the nth triangular numbers so we have to prove that this is the case. We use an algebraic approach for this one.
Note that this equation is called a recurrence relation or recurrence rule as it says how T(n) can be found recursively, step by step.
To see how this works we see that we are given T(1) = 1. Then we can get T(2) by substituting n = 2 in the first equation. So T(2) = T(1) + 2 = 3. Keep going to check out several more values of T(n).
Now that you are convinced that it works, start from the other end.
T(n) = T(n – 1) + n =[T(n – 2) + (n – 1)] + n = T(n – 2) + (n – 1) + n.
Successively repeating this process leads us finally to the expression for T(n) in (1). You should be able to see that this works for all values of n whether they are odd or even, so we don’t fall into the trap of (4).
Actually this last recurrence rule form of T(n) is best proved by a form of mathematical proof called
mathematical induction. This idea is too advanced for most students at this level but you may find a very bright student who can understand it.
Solution to the extension:
There are many other ways to show that these results hold or don’t hold. Let us know what you find and we’ll add it here with your school’s name attached.
Attachment  Size 

TriangularNumberLinks.pdf  58.19 KB 
TriangularNumberLinksMaori.pdf  50.68 KB 