# Mathematics Borders

Extracts from this document...

Introduction

Level 8

I will now express the generalization formula in terms of sigma notation using square numbers.

Below, I will show the sequence of cubes need to make a 3D cross shape through the help of level 7 in the form of a cross shape. This will be presented in a tabular manner

Shape | Number of cubes | Sigma Notation |

1 | 1 | 12+ 0 |

2 | 7 | 12+ 22 + 2 |

3 | 25 | 12 + 22 + 32 + 11 |

4 | 63 | 12+ 22 + 32 + 42 + 33 |

5 | 129 | 12+ 22 + 32 + 42 +52 + 74 |

6 | 231 | 12+ 22 + 32 + 42 + 52 + 62 +140 |

6Σ =r2

r = 1

Since I have found the formula

= Un = 4 n3 - 2n2 + 8 - 1

- 3

This

Middle

2nd difference → 5 7 9 11

3rd difference → 2 2 2

The 3rd difference is denoted by the term 6a.

Therefore in order to find a, I will replace the 3rd difference value

6a = 2

a = 2 = 1

6 3

By the given formula, Un = an3 + bn2 +cn + d, I will now be able to arrive to the generalized formula.

- If n=1

U1 = an3 + bn2 +cn + d

1 = 1 (1)3 + b(1)2 + c(1) + d

3

= b + c + d = 2 [equation 1]

3

- If n = 2

U2 = 1 (2)3 + b(2)2 + c(2) + d

3

5 = 8+ 4b + 2c + d

3

= 4b + 2c + d = 7 [equation 2]

3

- If n=3

U3 = 1 (3)3 + b(3)2 + c(3) + d

3

14 = 9 + 9b + 3c + d

= 9b + 3c + d = 5 [equation 3]

[equation 2 – equation 1]

4b + 2c + d = 7

3

b + c + d = 2

3

= 3b + c = 5/3 [equation3]

- [equation 3 – equation 2]

9b + 3c + d = 5

4b + 2c + d = 7

3

= 5b + c = 8 [equation 5]

3

- [equation 5 – equation 4]

5b + c = 8

3

3b + c = 5

3

= 2b = 1

= b = ½

Conclusion

- [equation 3 – equation 2]

9b + 3c + d = -16

4b + 2c + d = -6

= 5b + c = -10 [equation 5]

- [equation 5 – equation 4]

5b + c = -10

3b + c = -5

= 2b = -5

= b = -5

2

In order to find c I will replace the value b = -5 in equation 5

2

5 * -5+ c = -10

2

-25 + c = -10

2

c = -10 + 25

2

c = -20 + 25

2

c = 5

2

In order to find d I will now replace b and c in equation 1

-5+ 5 + d = - 1

2 2

0 + d = - 1

Therefore

a=1

b= -5

2

c=5

2

d= -1

Therefore generalized formula is = n3 – 5n2 + 5n – 1 [formula 2]

2 2

After obtaining both the formulas, I will now be adding formula 1 and formula 2 to justify the formula I got in level 7.

1n3 + 1n2 + 1n

3 2 6

1n3 – 5n2 + 5n – 1

2 2

= 4n3– 2n2 + 8n – 1

3 3

Thus I do obtain the formula which I had obtained in level 7. This thus justifies that my formula is right.

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

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