• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  • Level: GCSE
  • Subject: Maths
  • Word count: 2343

Borders Investigation

Extracts from this document...


Joseph Smith



The starting point for this investigation is the following sequence of cross shapes:


The sequence begins with a single white square, which is then surrounded by black squares to form the second shape. Each new cross is then formed by completely surrounding the previous cross with a border of black squares. In each new cross, the previous cross can be seen as the area of white squares in the centre.

The aim of this investigation is to derive algebraic formulae from the sequence, each expressing one property in terms of another (e.g. defining the area as the diameter squared). These formulae can then be examined and, hopefully, proven, using a wide variety of mathematical tools.

The next stage will then be to apply this same process to the three-dimensional counterpart of this sequence of shapes. This sequence would begin with a single white cube, which would then be surrounded on all six sides by black cubes, and the sequence would continue in the same way as the two-dimensional version.


Defining n

Throughout this investigation, the variable n (the value in the phrase “nth term”) is used to refer to terms in sequences. Terms are compared to, and expressed as functions of, n. However, in order to be able to identify the value of n

...read more.


i.e. remove the discrete square units) and then rotate it by 45°, we end up with a square, and what was previously the diameter across the middle of the cross is now the diagonal across the square.image06.png

We know that the diameter is equal to image30.png, and therefore this is the length of the diagonal across the square. Using this, we can derive a formula for the area of the square in terms of n.


Pythagoras’ theorem tells us that image35.png is equal to image36.png(because l forms a right-angled triangle with the square, and l is the hypotenuse while s is the length of both other sides), so we know that image37.png. As the area is also image38.png, we can say that the area equals image39.png. Now l, which is the diagonal across the square, is also the diameter of the cross, which as we have said is image30.png, so by substituting this into image40.png, multiplying out and simplifying, we get image41.png. This final result is, of course, image42.pngless than the actual formula, because when we round off the edges of the cross to remove the discrete units we also lose some accuracy.


Here we have our smooth approximation of the shape drawn over the top of the

...read more.


r does range from one to n.

So, we begin with a formula which we gives the volume of the nth 3D cross. It contains a series which is the sum of all those 2D cross layers above the middle. This uses the formula we have already obtained and proven for the area of a 2D cross, and, as we have said, substitutes in values ranging from zero to image65.png (because the number of these layers must be n, and the smallest 2D layer, because it is a single cube, must use 0 as the value of r). The sum of this series is multiplied by two (to include both above the middle and below it), and then the volume of the central layer (simply image53.png) is added. Written out, with sigma notation, it is:


What we must do next is convert the series so that r can range from one to n. Since the range 0 to image65.png is just the same as 1 to n shifted down by one, we can do this by replacing each r in the formula by image67.png. We can then substitute in standard results. There are three we must use now, one which we have already come across:


What follows, then, is our final proof:


The result is identical to our formula for volume. Both the 2D and 3D formulae have now been proven.

        -  -

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Fencing Problem essays

  1. Fencing investigation.

    S = half the perimeter (in this case 500) V (S (S-a)(S-b)(S-c)) V500 (500-350)(500-200)(500-450) V500 x 150 x 300 x 50 V1125000000 33541.02m2 Review In the previous section: I have noticed that the more regular the triangle is, the larger its area will be.

  2. Perimeter Investigation

    In a regular pentagon each equal side would be = 1000 / 5 = 200m Using TAN = h / 100 = tan 54 h = 100tan54 = 137.6 Area = 1/2 * 200 * 137.6 = 13760 Area of 5 triangles = 13760 * 5 = 68 800m In

  1. Geography Investigation: Residential Areas

    or privately rent because I am suspecting that there will be a higher amount of private rented properties because people don't want to spend the rest of their lives living in the inner city. 3) Externalities will have a higher penalty point than the outskirts of the town.

  2. Biological Individual Investigation What Effects Have Management Had On Grasses In Rushey Plain, Epping ...

    This meant that light was the key variable in affecting grass height. The top 7cm was the humus layer, which was dark brown in colour, and was quite dry. The 7cm - 18cm layer was a silt loam layer, with a light brown colour, and was slightly moist.

  1. Geography As Environmental Investigation

    Many people may be too busy to answer the questions on the questionnaire. Traffic flow Used to find out number of vehicles in usage at each site On each of the sites I will count the number of vehicles in usage.

  2. Regeneration has had a positive impact on the Sutton Harbour area - its environment, ...

    When I found a good place, I took the picture. The picture of the 'Coxside Leisure Park' signpost is shown below. Sketches Sketches are much like the photographs. I chose to include a couple of hand drawn images so that it varied my sources of information.

  1. Beyond Pythagoras

    Difference 1 8-4 4 4 2 24-16 8 4 3 48-36 12 4 4 80-64 16 4 5 120-100 20 4 Rule = 4n So the rule that the adjacent line column follows is = 4n�+4n This is how I found out the rule the third column follows (hypotenuse line)

  2. Maths Investigation on Trays.

    we have the highest volume of the turquoise line. x Area Base Area 4 sides Volume 2.2 184.96 119.68 406.91 2.4 174.24 126.72 418.18 2.6 163.84 133.12 425.98 2.8 153.76 138.88 430.53 3 144 144 432 3.2 134.56 148.48 430.59 3.4 125.44 152.32 426.5 3.6 116.64 155.52 419.9 3.8 108.16 158.08 411.01 4 100 160 400 Here we can

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work