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• Level: GCSE
• Subject: Maths
• Word count: 1078

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Introduction

Mathematic Coursework

Hossay Rahimy 11 Fisher

Introduction

In this coursework I am going to investigate to see how many squares would be needed to make any cross-shape built up in this way.

▬►                                     ▬►

We can see that number of black is the total number of the pattern before.

A

 Pattern Black White Total 1 1 4 5 2 5 8 13 3 13 12 25 4 25 16 41 5 41 20 61 6 61 24 85

White

To get the formula for the a quadratic sequence is: An²+ B+ C

I have to find a formula for the nth term of this number sequence:

4, 8, 12, 16, 24,

 White 4 8 12 16 20 24 Difference 4 4 4 N term 4N

The first difference is a constant for the white pattern, so the nth term is:  4N

Black

I have to find a formula for the nth term of this sequence:

1, 5, 13, 25, 41, 61,

Black:                                                                 1       5       13       25      41      61

\  /     \  /     \   /    \    /    \   /

First differences are not the same:             4        8        12       16       20

\   /    \    /    \    /   \    /

Second differences are the same:                        4         4       4         4

Formulas that contain n²: If the second difference is a constant,                   the formula for the nth term contains n². The number in front of n² is

Middle

-1

-3

-5

-7

-9

-11

\              /  \             /      \         /       \        /        \           /

-2                   -2                -2                 -2                   -2

The formula for the *Rest of sequence* is -2n+1

(Since it is multiples of -2, plus 1)

The final formula is 2n²-2n+1

I will check the formula by finding term number 7.

n=7, 2n² -2n+1= 85

85 should fit the difference pattern at the start of the sequence.

Total

I have to find a formula for the nth term of this number sequence:

5, 13, 25, 41, 61, 85

Total                                                                  5      13       25       41      61       85

\  /     \    /     \   /     \  /    \    /

First differences are not the same:             8          12       16       20      24

\  /       \   /    \   /    \    /

Second differences are the same:                       4           4        4        4

Formulas that contain n²: If the second difference is a constant,                   the formula for the nth term contains n². The number in front of n² is half the constant difference.

The constant difference is 4.

The number in front of n² is half of 4, which is 2.

The first part of the formula is therefore 2n².

Line 1:    the number of each term

Line 2:    the number is the sequence

Line 3:     2n² worked out from the number of each term

Line 4:    the rest of sequence comes from taking 2n² away from the numbers in the sequence (line 2 take away line 3)

 1.number of term(n) 1 2 3 4 5 6 2.sequence 5 13 25 41 61 85 3. 2n² 2 8 18 32 50 72 4.rest of sequence 3 5 7 9 11 13 \              /  \             /      \         /       \        /        \           /2                  2                2                 2                   2

Conclusion

p.s. 4/3 means  4 over 3

 1. number of n term 1 2 3 4 5 2.Sequence 7 25 63 129 231 3.  4n³/3 4/3 32/3 108/3 256/3 500/3 4. rest of sequence 21-4/3 75-32/3 189-108/3 387-256/3 693-500/3 \             /          \               /    \               /             \        /              26/3                   38/3            50/3                    62/3                       \               /           \            /    \                   /                            4                            4                 4

Formulas that contain n²: If the second difference is a constant,                   the formula for the nth term contains n². The number in front of n² is half the constant difference.

The constant difference is 4.

The number in front of n² is half of 4, which is 2.

The first part of the formula is therefore 2n².

The rest of sequence of a cubic formula is the sequence for the quadratic formula, so the sequences are:

5 2/3, 14 1/3, 27, 43 2/3, 64 1/3

 1.numberof tern (n) 1 2 3 4 5 2. sequence 5 2/3 14 1/3 27 43 2/3 64 1/3 3.2n² 2 8 18 32 50 4. rest of sequence 3 2/3 6 1/3 9 11 2/3 14 1/3 \                  /             \           /        \         /             \                  /                                     2 2/3                        2 2/3             2 2/3                2 2/3

The formula for the *Rest of sequence* is 2 2/3n+1

(Since it is multiples of 2 2/3, plus 1)

The final formula is 4n³/3+2n²+8n/3+1

I will check the formula by finding term number 5.

n=5, 4/3n³+6/3n² +8/3n+1= 231

231 fit in pattern of the sequence.

-  -

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