Algebra with Multilink CubesAuthor(s): Steve AbbottSource: Mathematics in School, Vol. 21, No. 2 (Mar., 1992), pp. 12-13Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30214849 .

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by Steve Abbott Farlingaye High School, Suffolk

In a recent series of lessons with 8A, a mixed ability group, we explored some sequences of patterns with Multilink cubes. Pupils had a free choice of which pattern to explore, but were given guidance on the likely difficulty. The patterns investigated ranged from a simple sequence of crosses: shown in Figure 1 to the triangular pyramid de-

Fig. 1

scribed below by Victoria Garfath-Cox. One pair, Tracy Hempel and Tina Smith, investigated the square pyramid corresponding to the sum

12 + 22 + 32 + 42 +...

They had seen in an earlier lesson that two staircases can be put together to form a rectangle, as explained below by Victoria, and so experimented with fitting together their square pyramids. They found a consistent way of fitting three pyramids, and with some help were able to make cuboids by joining two copies of their three-pyramid shapes. Working with Victoria, they then found the formula referred to below.

What was successful about these lessons? From the pupils point of view, there was an interesting problem to solve. The problems were posed by the pupils themselves and were sufficiently non-trivial to allow useful group discussion. Each pupil was expected to write up the group results, so there was an incentive for each member of a group to really try to understand the problem and its solution. From the teacher's point of view, the pupils were engaged on tasks which required them to use algebraic ideas, to generalise, and which brought home the need for some kind of symbolic notation. Above all, there was an opportunity to work at a concrete level, using the cubes to test conjectures. The use of cubes meant that there were three possible strategies- geometric, numerical and algebraic. Victoria's work below involves a synthesis of these strategies, and includes one 'borrowed' result.

One of the difficulties encountered at A-level is the so- called 'algebra gap'. In fact, the gap is more to do with

12 Mathematics in School, March 1992

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manipulation skills than the understanding of algebraic concepts. It is possible to build activities, like the cubes investigation, into lessons throughout the 11-16 phase. If given suitable encouragement, those pupils who are likely to need manipulation skills will develop them, slowly, and as the need arises. Many of the reasons put forward for the inclusion of mathematics in the curriculum for all pupils, concern mathematical processes rather than particu- lar content. We should choose our content to fit the processes. Our more able pupils can be guided to develop certain skills further, but it surely helps them if they have met several contexts in which the skills are useful.

The remainder of this note is Victoria's account of her investigation. Apart from general guidance, she needed help only with the simplification of the final formula. I suspect that on a future occasion, Victoria will want to improve her skills in algebraic manipulation, so that her solution will be all her own work.

Cubes (Victoria's account) My task: I chose a shape made out of cubes and I had to find out how many cubes there were in this shape. I had to find a way of doing this without counting every cube, however many layers there are. It was difficult because there are a different number of cubes in each layer. The shape looked like this:

Fig. 2

I found that if I knew the number of the layer, (e.g. the fourth layer or the fifth layer) I could find out how many cubes there were in it. I did this by taking the layer number, n, and multiplying by itself plus one and halving it. This is because if you put two layers together they make a rectangle like this:

Fig. 3

The formula for this is 1/2n(n+1), which is the same as

1/2(n2 + n).

So you can find the cubes in the pyramid by

1/2(12+ 1) +1/2(22+ 2) +1/2(32+ 3) + ...

But this would take a very long time to do, especi- ally if you had a shape with 100 layers, so I tried to make it simpler:

1/2(12+1) +1/2(22 + 2) etc.

is the same as

1/2(12 + 22 + 32 + .--) + 1/2(1 + 2 + 3 + .--). I started with the last bit of the sum:

(1 + 2 + 3 +)

which is just the staircase again,

1 +2

+3 +4

Fig. 4

so I can use the formula 1/2n(n+ 1). The next bit was harder, but I knew that it was

the same as one of the problems that another group was doing. They were trying to find out how many cubes there were in a shape that had squares of cubes piled on top of each other. They had found that 6 shapes fitted together made a cuboid. (This is difficult to show in a diagram. The best way to see this is to make 6 copies of Fig. 5 from Multilink cubes, and then experiment with them.) There was a formula for finding the area of this cuboid, which was the number of the layer multiplied by that number plus one multiplied by the sum of those two numbers, e.g. 2 x 3 x 5, or

n(n+1l)(2n+l). This is for six shapes, so to find one I had to divide this number by six.

Fig. 5

So I had found how to deal with each part of the problem. Next I had to join the two together. This gives:

1/2n(n+ 1) + n(n+l) (2n+ 1)/6.

Then the whole lot has to be divided by two because it was all halves to begin with, and the final formula looks like this:

1/2{1/2 n(n+ 1) + 1/6n (n+ 1)(2n+ 1)). This can be simplified to:

1/6 n(n+ 1)(n+ 2).

So for 100 layers, for example, you need

1/6 x 100 x 101 x 102 = 171700 cubes.

Mathematics in School, March 1992 13

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