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Profilirana Prirodomatematicheska Gimnaziya “Akademik Nikola Obreshkov”
ERASMUS+ PROJECT " DITUM'-Discover the Unknown in Mathematics’ Nr FI01-KA _5 THIS PROJECT HAS BEEN FUNDED WITH SUPPORT FROM THE EUROPEAN COMMISSION. THIS PUBLICATION [COMMUNICATION] REFLECTS THE VIEWS ONLY OF THE AUTHOR, AND THE COMMISSION CANNOT BE HELD RESPONSIBLE FOR ANY USE WHICH MAY BE MADE OF THE INFORMATION CONTAINED THEREIN.
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Тhe remarkable Pascal’s triangle
DITUM Bulgarian team presentation
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Blaise Pascal Mathematician Blaise Pascal was born on June 19, 1623, in Clermont-Ferrand, France. In the 1640s he invented the Pascaline, an early calculator, and further validated Evangelista Torricelli's theory concerning the cause of barometrical variations. He also laid the foundation for the modern theory of probabilities.
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1 2 1 3 4 6 5 10 15 20 1 7 21 35 8 28 56 70 Mark 1 The natural numbers can be found ascending along the second diagonal of the triangle.
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The number of dots needed to form an equilateral triangle is a triangle number.
= = = = = =28
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Mark 2 The triangle numbers are to be found ascending along the third diagonal of the triangle. Triangle numbers are numbers that one can find with the following formula: 1 2 1 3 4 6 5 10 15 20 7 21 35 8 28 56 70
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Interesting fact: Each square can be presented as an addition of two diagonally-neighbouring triangle numbers.
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Mark 3: Square numbers * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * 1 * * * * * * * * * * * * * * 4 * * * * * * * * * * * * * * * * * * * * * * * * * * 25
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Mark 4 Hexagonial numbers are every second number along the third diagonal. Their formula is 1 2 1 3 4 6 5 10 15 20 7 21 35 8 28 56 70
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Mark 5 Fibonacci numbers start with one and every following one is a sum of the previous two (1+2=3). Such numbers are: 1, 1, 2, 3, 5, 8, 13, 21. To find them you need to go at an angle along the shallow diagonals. xn = xn-1 + xn-2
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Task: Find the standart form of :
(x + y)2 = (x + y)3 = (x + y)4 = x + y x + y x2 + xy xy + y 1x2 + 2xy +1y2 x2 + 2xy + y2 x + y x3 + 2x2y + xy2 x2y + 2xy2 + y x3 + 3x2y + 3xy2 + 1y3 x3 + 3x2y + 3xy2 + y3 x + y x4 + 3x3y + 3x2y2 + xy3 x3y + 3x2y2 + 3xy3 + y x4 + 4x3y + 6x2y2 + 4xy3 +1y4 In order to define the coefficients in the standard form of the polynomial 𝒙+𝒚 𝒏 we examine the numbers in the n-th row of Pascal’s triangle. (given first row as nil).
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Binomial coefficients
Mark 6 Binomial coefficients 1 1x + 1y 1x2 2xy 1y2 1x3 3x2y 3xy2 1y3 1x4 4x3y 6x2y2 4xy3 1y4 1x5 5x4y 10x3y2 10x2y3 5xy4 1y5
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II. 11 10 we examine 𝟏𝟎 𝒕𝒉 row in Pascal’s triangle:
Mark 7: Powers of 11. I. Using Mark 6: 115 = 1(105) + 5(104) + 10(103) + 10(102) + 5(101) + 1(100) = = = = II. we examine 𝟏𝟎 𝒕𝒉 row in Pascal’s triangle: 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. 01 10 45 120 210 252 ______________ = 𝟏𝟏 𝟏𝟎
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Mark 8: Powers of 2 The 𝒏 𝒕𝒉 power of 2 in Pascal’s triangle is
1 1 2 2 4 3 8 4 6 16 5 10 32 15 20 64 7 21 35 128 8 28 56 70 20 = 1 21 = 2 22 = 4 23 = 8 24 = = = 64 27 = 128 28 = 256 The 𝒏 𝒕𝒉 power of 2 in Pascal’s triangle is the sum of numbers along the 𝒏 𝒕𝒉 row
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Mark 9: Symmetry in Pascal’s triangle.
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Thank you for the attention
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