1. 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.

2. Тhe remarkable Pascal’s triangle
DITUM
Bulgarian team presentation

3. 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.

4.  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.

5. The number of dots needed to form an equilateral triangle is a triangle number.
= = = = = =28

6. 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

7. Interesting fact: Each square can be presented as an addition of two diagonally-neighbouring triangle numbers.

8. Mark 3: Square numbers * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * *
1 * * * * * * * * * * * * * *
4 * * * * * * * * * * * *
* * * * * * * * *
* * * * * 25

9. 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

10. 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

11. 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).

12. 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

13. 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

14. 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

15. Mark 9: Symmetry in Pascal’s triangle.

16. Thank you for the attention