• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Triminoes Investigation

Extracts from this document...

Introduction

G.C.S.E Mathematic Coursework                                              Triminoes Investigation

Triminoes Investigation

Sequences

There are different lots of sequences such as Arithmetic progressions, Geometric progressions and other sequences. There are many types of sequences, sequences that increase by a fixed amount between each term are known as arithmetic progressions or arithmetic series. Odd and even numbers both increase by two from one term to the next. The multiples of seven increases by seven, from one term to the next.

Arithmetic progressions

There are many types of sequences, sequences that increase by a fixed amount between each term are known as arithmetic progressions or arithmetic series. Odd and even numbers both increase by two from one term to the next. The multiples of seven increases by seven from one term to the next.

Odd numbers         1, 3, 5, 7, 9,   …
Even numbers
2, 4, 6, 8, 10, …
Multiples of seven
7, 14, 21, 28, … are examples of arithmetic progressions or arithmetic series.

### Geometric progressions

Geometric progressions or geometric series are sequences in which successive terms are in the same ratio. To get the next power of two you simply double the previous value.
In the example of exponential growth the next term is obtained by multiplying the previous term by
1.5.

Powers of two          1, 2, 4, 8, 16, …
Exponential growth 10, 15, 22.5, 33.75, …
Exponential decay   20, 16, 12.8, 10.24, 8.192, …
are examples of

Middle

-        19a + 5b + c = 10

18a + 2b       =  5 – Equation 8

Equation 6 – Equation 7

I am doing this to eliminate c from the two equations, to create an equation.

19a  + 5b  + c  = 10   – equation 6

-    7a  + 3b  + c  =  6    – equation 7

12a  + 2b         = 4     – equation 9

Equation 8 – Equation 9

I am doing this to cancel out b and find the value of a.

18a + 2b = 5

12a + 2b = 4

6a         = 1

a = 1/6

I am going to Substitute a = 1/6 into equation .

I’m doing this because to find what “b” is worth when substituting a = 1/6 into equation 9.

18a + 2b        = 5

18 x 1/6 + 2b = 5

3 + 2b = 5

2b = 5 - 3

2b = 2

b = 2/2

b = 1

Substitute a = 1/6, b = 1 into equation 6.

I’m doing this because to find what “c” is worth when substituting a = 2, b = 3 into equation 6.

19a + 5b + c            = 10

19 x 1/6 + 5 x 1 + c = 10

19/6 + 5 + c             = 10

3 + 1/6 + 5 + c         = 10

8 + 1/6 c   = 10

c    = 10 – 8 - 1/6

c    = 11/6

I am going to Substitute a = 1/6, b = 1 and c = 11/6 into equation 1.

I’m doing this because to find what “d” is worth and also that leads me to working out the formula for the area of the triangle.

a + b + c + d         = 4

1/6 + 1 + 11/6 + d = 4

12/6 + 1 + d = 4

2 + 1 d = 4

d = 4 –3

d = 1

I am going to Substitute a = 1/6, b = 1, c =11/6 and d = 1 put into the quadratic sequence.

F (n) = an³ + bn² + cn + d

F (n) = 1/6n³ + 1 x n² + 11/6n + 1 Conclusion

I am going to find the value of the negative number so I could see how it goes in the graph.

F(n) = 1/4n (n³ + 6n² + 11n + 6)                F(-1) = 1/4n ((-1)³ + 6 x (–1)² + 11 x -1 + 6)

F(0) = 1/4n (0³ + 6 x 0² + 11 x 0 + 6)                              = 1/4n ( -1 + 6 + -11 + 6)

= ¼ x 0 x 6                                             = ¼ x –1 (0)

= 0                                                     = - ¼ x 0 = 0

F(-2) = 1/4n ((-2)³ + 6 x (–2)² + 11 x -2 + 6)   F(-3) = 1/4n ((-3)³ + 6 x (–3)² + 11 x -3 + 6)

= 1/4n ( -8 + 24 –22 + 6)                          = 1/4n ( -27 + 54 –33 + 6)

= ¼ x –2 x 0                                           = ¼ x –3 x 0

= 0                                                    = 0

F(-4) = 1/4n ((-4)³ + 6 x (–4)² + 11 x -4 + 6)   F(-5) = 1/4n ((-5)³ + 6 x (–5)² + 11 x -5 + 6)

= 1/4n (-64 + 96 – 44 + 6)                          = 1/4n (-125 + 150 –55 + 6)

= ¼ x – 4 x - 6                                           = ¼ x –5 x -24

=  24/4 = 6                                          = 120/4 = 30

F(-6) = 1/4n ((-6)³ + 6 x (–6)² + 11 x -6 + 6)

= 1/4n (-216 + 216 – 66 + 6)

= ¼ x – 6 x – 60

=  360/4 = 90

Using the results I have created a table which will be much easier to see and to plot the points in the graph “f(n) = 1/4n (n³ + 6n² + 11n + 6)”.

 X -6 -5 -4 -3 -2 -1 0 1 2 3 Y 90 30 6 0 0 0 0 6 30 90

Now I have to draw a graph for both formulas:-

1. 1/6 (n³ + 6n² + 11n + 6)
2. 1/4n (n³ + 6n² + 11n + 6) This graph is for the Cubic formula. This graph is for the Quartic formula.

Conclusion

The relationship:

If you could see

1. 1/6 (n³ + 6n² + 11n + 6)
1. 1/4n (n³ + 6n² + 11n + 6)

that both of the formulas ends up with the same numbers and letters inside the brackets.

Abdal MOHAMED 11.4

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics 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

# Related AS and A Level Core & Pure Mathematics essays

1. ## Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

Method of Bisection: n a f(a) b f(b) c f(c) error 1 0 1 1 -0.5 0.5 0.3125 0.5 2 0.5 0.3125 1 -0.5 0.75 -0.1015625 0.25 3 0.5 0.3125 0.75 -0.1015625 0.625 0.106445313 0.125 4 0.625 0.106445313 0.75 -0.1015625 0.6875 0.002319336 0.0625 5 0.6875 0.002319336 0.75 -0.1015625 0.71875 -0.049697876

2. ## Functions. Mappings transform one set of numbers into another set of numbers. We could ...

3. To start, substitute x0 Different rearrangements of f(x) may produce different roots x0 needs to be close to the root Sometimes the sequence won't converge to a root Modulus Function The modulus of a number a is written as |a| and is read mod a It is its positive numerical value The modulus function of f(x)

1. ## Functions Coursework - A2 Maths

This occurs because although the graph crosses three times, two of those crossings are in the same interval on the table. A change of sign occurs when the graph crosses the x-axis. This graph crosses the x-axis twice in the interval [-2,-2].

2. ## Sequence &amp;amp; Series

is twice the 5th term. Show that the common difference is equal to the first term. From the question we know that, where and. Therefore,. 10. Three numbers in an A.P have sum 33 and product 1232. Find the numbers. Let the numbers be a, b and c.

1. This would indicate that it has only one root within the interval and the second root is over looked. I will use the equation 0 = x4+5x3-7x+2 to demonstrate how this method would fail. Here is a graph of the function f(x)

2. ## Solutions of equations

So taking the answer as the midpoint of the two I get: x = 0.666665 (6 d.p.). But seeing as I only need the answer to 5 decimal places I get x = 0.66667 (5 d.p.). Answer: 0.66667 (5 d.p.)

1. ## Mathematical Investigation

Part II Investigate the graphs of y=sin (b�x) Figure #4: Comparison of graphs of y=sin(x); y=sin (2x) and y=sin (4x). According to Figure#4, as the value of "b" increases from 1 to 2, as indicated by the red graph [y=sin(x)] and blue graph [y=sin(2x)], one complete cycle of the blue graph (1pi radians)

2. ## Fractals. In order to create a fractal, you will need to be acquainted ...

The iterations tend to get really big; however, you can find the outputs of every iteration over here: http://www.michaelnorris.info/software/l-system-generator.html . On the computer, you can simulate the creation of fractals using the L-system by assigning these rules to a turtle-based programming language (logo programs, Netlogo) • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 