Patterns in Context Part D: Counting Stairs (60 minutes)

Note 10

Think about the different conceptions of algebra and the use of variables when working on these problems.

For those not using the Counting Stairs interactive activity, each individual working alone or group should have about 30 cubes to use.

Whether doing the Counting Stairs interactive activity or using blocks or cubes, come up with as many ways as possible for solving the problem. (There are at least 12 different solutions.) If you get stuck, think about building squares or rectangles from the stairs. In fact, the geometric solution to this problem is the most elegant: If you take any staircase with bottom row length n, form a second identical one, rotate it, and put it together with the first, you form a rectangle of dimensions n by n + 1. Therefore, the area of the original staircase is 1/2(n)(n+1).

Groups: Share answers to Problem D1. If anyone used variables, they should describe what conception of variable they were using.

Note 11

Groups: It is important to talk about Problem D2 as a whole group, as this question underscores the importance and convenience of the recursive formula. If you know the number of stairs in the nth staircase, the number of stairs in the next staircase can be found by adding n + 1 onto the previous total.

Notice that the recursive formula can be applied only in situations where you know the previous term. Part of our job is to determine in which situations different kinds of representations are most useful. What makes a rule useful is how easy it is to apply — closed-form rules often win the battle here — and how easy it is to come up with. In the staircase problem, the recursive rule is much easier to come up with than the closed form.

Another example where the recursive rule is easier to find than the closed-form rule is the famous Fibonacci number sequence: 1, 1, 2, 3, 5, 8, 13, … .

The recursive rule for the sequence is elegant: Start with 1, 1. Then, to get any term in the sequence, add the previous two terms.

The closed-form rule is complicated:

It’s not clear that this function produces integer outputs, much less that the outputs are the terms of the sequence above. Imagine trying to come up with such a formula!

Problem D1

One method is that the number of blocks for the nth staircase is n more than the number of blocks for the (n – 1)st staircase, because you need to add on a row of n blocks to build the new staircase. This allows you to construct a table. One clever trick is to piece together 2 consecutive staircases; the 5th staircase fits nicely into the 6th staircase to form a 6-by-6 square!

Problem D2

There will be 1 new row in staircase n + 1, a row with n + 1 blocks in it. That means that to go from staircase n to staircase n + 1, you will need to add n + 1 blocks.

Problem D3

In the 274th staircase there will be 37,401 + 274 = 37,675 blocks. In the 275th staircase there will be 37,675 + 275 = 37,950 blocks. This answer can also be reached by using (275) * (276) / 2 = 37,950.

Problem D4

There are 5,050 blocks in the 100th staircase. The fastest way is to use n(n + 1) / 2.

Problem D5

A way to build a general formula is to notice that 2 copies of the 6th staircase will form a 6-by-7 rectangle. For the nth staircase this will be n(n + 1) blocks for 2 staircases, which means n(n + 1) / 2 for 1.