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By: Michelle Green. Construction of Pascal’s Triangle 1 0+1 1+1 4 1+2 0+1 41 6 1 10 55 1 1 ROW 0 ROW 1 ROW 2 ROW 3 ROW 4 ROW 5 20151 1 ROW 6 ALWAYS.

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Presentation on theme: "By: Michelle Green. Construction of Pascal’s Triangle 1 0+1 1+1 4 1+2 0+1 41 6 1 10 55 1 1 ROW 0 ROW 1 ROW 2 ROW 3 ROW 4 ROW 5 20151 1 ROW 6 ALWAYS."— Presentation transcript:

1 By: Michelle Green

2

3 Construction of Pascal’s Triangle 1 0+1 1+1 4 1+2 0+1 41 6 1 10 55 1 1 ROW 0 ROW 1 ROW 2 ROW 3 ROW 4 ROW 5 20151 1 ROW 6 ALWAYS start with a 1 To fill in a number, add the numbers above it to the left and right. 0 + +0 11 0+1 0+ +0 121 1331 6 206

4 Patterns Within Pascal’s Triangle Horizontal Sums Prime Numbers Hockey Stick Exponents of 11

5 Horizontal Sums 1 0+1 1+1 4 1+2 0+1 41 6 1 10 55 1 1 ROW 0 = 1 ROW 1 = 1+1=2 ROW 2 = 1+2+1=4 ROW 3 = 1+3+3+1=8 ROW 4 = 1+4+6+4+1=16 2015 1 1 11 0+1 121 1331 6 206 ROW 5= 1+5+10+10+5+1= 32 ROW 6= 1+6+15+20+20+6+1= 64 Can you see the pattern of the horizontal sums? BACK

6 Prime Numbers For any row that begins (not counting the one) with a prime number, all other numbers are divisible by the prime number. Examples: – Row 5: 1 5 10 10 5 1 both 5 and 10 are divisible by 5 – Row 11: 1 11 55 165 330 462 462 330 165 55 11 1 11, 55, 165, 330, 462 are all divisible by 11 BACK

7 Hockey Stick Pattern 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 BACK

8 Exponents of 11 BACK 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 1 11⁰=1 11 ¹ =11 11 ² =121 11 ⁵ =?? 16105115 10 1051

9 Using Pascal’s Triangle Pascal’s Triangle can be used to expand binomials – (a+b)ⁿ For example we can use Pascal’s Triangle to expand: (x+3) ² –Since we know how to FOIL, we know the answer is x*x+3x+3x+9=x²+6x+9 – All we need to do is take a row from Pascal’s Triangle Always use the row of the exponent (here we need row 2 of Pascal’s Triangle)

10 Expanding (x+3) ² Take row 2 of Pascal’s Triangle – 1,2,1 – These will be our coefficients Starting with 2, decrease the exponent for x. Also, start with 0 and increase the exponent for 3. 1 (x) ²(3)º + 2(x)¹(3)¹ + 1(x)º(3)² x²+6x+9 Using the triangle may seem harder, but for higher exponent, it is easier than multiplying everything out.

11 Expand (2x+6)⁵ This binomial will be too time consuming to multiply out, so we will use Pascal’s Triangle. Take row 5 of Pascal’s Triangle – 1 5 10 10 5 1 1(2x)⁵(6)⁰ + 5(2x)⁴(6)¹ + 10(2x)³(6)² + 10(2x)²(6) ³+ 5(2x)¹(6)⁴ + 1(2x)⁰(6)⁵ 1(32x⁵)1 + 5(16x⁴)6 + 10(8x³)36 + 10(4x²)(216) + 5(2x)(1296) + 1(1)(7776) 32x⁵ + 480x⁴ + 2880x³ + 8640x² + 12960x + 7776

12 Practice Using your Pascal’s Triangle, expand the following binomials: – (x+4)² – (5x+2)³ – (3x+1)³ – (2x+9)⁴

13 Resources Pascal’s Triangle and Its Patterns http://ptri1.tripod.com/#hockey. 16 October 2011.


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