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# Maths Investigation: Pascals Triangles

Extracts from this document...

Introduction

Maths investigation

John Abarshi

Part 1: Rows Pattern

From the first 5 rows, I observe a pattern which is repeated on the subsequent row. First, the number 1 stands at the pivot of the pyramid and begins and ends each row. Second, the numbers follow a sequence 1,2,3,….both diagonally and downwards, and are same per row on the right and left sides of the pyramid, and the first number coincides with the number of the row. A similar sequence follows next (1,3,6..) on both sides, and then (1,4,10…) etc. Third, every number on each row (excluding the 1’s) is the sum of the two numbers sited immediately above it. Four, on every other row (i.e. the even number rows), there is a centrally-located number (2,6,20..), and this also has a sequence; it corresponds to the sum of the same number (1+1,3+3,10+10). The rows that fall in between consist of and even number of digits (i.e. no middle number here), but the two middle numbers are always the same (1,1; 3,3; 10,10;..). Therefore you have a

Middle

4        1

1        5        10        10        5        1

1         6         15        20        15        6        1

1        7        21        35        35        21         7        1

1        8        28        56        70        56        28        8        1

1        9        36        84        126        126        84        36        9        1

1        10        45        120        210        252        210        120        45        10        1

1        11        55        165        330        462        462        330        165        55        11        1

1

1        1

1        2        1

1        3        3        1

1        4        6        4        1

1        5        10        10        5        1

1         6         15        20        15        6        1

1        7        21        35        35        21         7        1

1        8        28        56        70        56        28        8        1

1        9        36        84        126        126        84        36        9        1

1        10        45        120        210        252        210        120        45        10        1

On shading multiples of 3 (pink), there were inverted triangles that formed a pattern around the middle of the pyramid. The smaller triangles stood on a base of 2 numbers, and the base of two triangles was separated by one number (20). The height of the triangle is also two rows.

On shading multiples of 5, we have an inverted triangle which again is symmetrical around the middle on a base of four numbers. This pattern is then repeated but this time on two sides of the pyramid, and the base is again separated by one number (252). In this case, the height of the triangle is four rows.

Explanation: Perhaps the triangular patterns are always formed on a base and height of (n-1) in a symmetrical pattern, like two sides of an equilateral triangle.

Part 4: Pascal Petals

1.

1

1        1

1        2        1

1        3        3        1

1        4        6        4        1

1        5        10        10        5        1

1         6         15        20        15        6        1

1        7        21        35        35        21         7        1

1        8        28        56        70        56        28        8        1

1        9        36        84        126        126        84        36        9        1

1        10        45        120        210        252        210        120        45        10        1

Cell 21 (grey petal):

6x28x35 = 5880 = (2x3)x(2x2x7)x(5x7)

15x7x56 = 5880 = (3x5)x(1x7)x(2x2x2x7)

Conclusion

3rd triangular number: is 6 and then you add 4.

4th triangular number:  it will be 10 and you add 5 this time.

5th triangular number: will be 15 and after this you have to add 6.

And so on so on. This pattern will stay for the rest of Pascal’s triangle.

 Row 4 Row 3 Row 2 Row1 N1 1 +3 N2 4 +3 +6 +1 N3 10 +4 +10 +1 N4 20 +5 +15 +1 N5 35 +6 +21 N6 56

Extra:

1. (a+b)6

Since (a+b)3 = a3+3a2b+3ab2+b3  i.e. row 3 pattern of ……        1        3        3        1

And row 6 pattern is….1        6        15        20        15        6        1

Therefore, (a+b)6 = a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6

1. (a-b)7

Since row 7 pattern is….1        7        21        35        35        21        7        1

Therefore, (a-b)7 = a7-7a6b+21a5b2-35a4b3+35a3b4-21a2b5+7ab6-b7

1. (2x-1)3

And (a+b)3 = a3+3a2b+3ab2+b3; where a=2x and b=-1

Answer: (2x)3 + 3.4x2.-1 + 3.2x.1 + (-1)3  = 8x3- 12 x2+ 6x.1 -1.

1. (x+1/x)4

Since row 7 pattern is….1        7        21        35        35        21        7        1

And (a+b)4 = a4+4a3b+6a2b2+4ab3+b4; where a=x and b=1/x

Answer: x4+ 4x3/x + 6x2/x2+ 4x/x3+1/x4  = x4+ 4x2 + 6 + 4/x2+1/x4

= x4+ 4x2 + 6 + 4/x2+1/x4

Bibliography

1. http://mathforum.org/workshops/usi/pascal/elem.color_pascal.html

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