Patterns & Math in Pascal s Triangle!

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1 Uses for Pascal s Triangle An entry in Pascal's Triangle is usually given a row number and a place in that row, beginning with row zero and place, or element, zero. For instance, at the top of Pascal's Triangle is the number 1, which makes up row zero, element zero. The next row, row 1, contains two 1's, both formed by adding the two numbers above them to the left and the right, in this case 1 and 0 (all numbers outside the Triangle are 0's). The 2nd row is created as follows: 0+1=1; 1+1=2; 1+0=1. The third is: 0+1=1; 1+2=3; 2+1=3; 1+0=1. Likewise, the number 20 appears in row 6, element 3 in Pascal s Triangle. In this way, the rows of the triangle go on infinitely. Patterns & Math in Pascal s Triangle! The Sums of the Rows Put a token on each of the elements in any one row. Take the sum of the numbers on that row and write the total as a power of 2. What do you find? (They will be equal to 2 to the n th power or 2 n, when n is the number of the row). For example if you cover all elements of the 6 th row, then add them up, you would get: = 64 = o = = = = = = = = = 16 Written by Jeanne Lazzarini (RAFT)

2 Prime Numbers If the 1 st element in a row is a prime number then cover that number with a token (remember, the zero th element of every row is 1). What do you notice about all the other numbers in that same row (excluding the 1's)? You should find that all the numbers are divisible by that prime number: for example, in row 7; 7, 21, and 35 are all divisible by 7. Find Any Number in the Triangle! Any number in the triangle can also be found by ncr (n Choose r) where n is the number of the row and r is the element in that row. For example, in row 3, 1 is the zero th element, 3 is the 1 st element, the next 3 is the 2 nd element, and the last 1 is the 3 rd element. The formula for ncr is: n! r!(n-r)! where! means factorial, or the number multiplied by all the positive integers that are smaller than the number. For example: 5! = = 120. For example, ask a friend to find your number on the board by giving them only the number of the row (in this case 7) and the element number where 35 appears in that row (which is 3). Using the formula above, they should find (7!) [(3!) (7 3)]! = (7 x 6 x 5 x 4 x 3 x 2 x 1) [(3 x 2 x 1) (4 x 3 x 2 x 1)] = 7 x 5 = 35; so 35 was the number!! The Probability of Heads or Tails Pascal originally discovered the properties of this triangle by thinking of problems posed by gamblers. Investigate what happens when a coin is flipped. If you flip a coin, it will come up either heads or tails. In the following analysis the focus is on the probability of flipping heads. An equivalent could be said for flipping tails. Cover the first 1 at the top of Pascal s Triangle (row 0) with a token to represent the flip of zero coins. Next, cover the numbers on the 1 st row to represent the probability of getting a head on the toss of 1 coin. There are only 2 possibilities when tossing one coin and the probability of the coin coming up heads is 1 chance out of 2, or ½. The probability of any event is expressed as a fraction between 0 and 1 inclusive. Compare the number equivalent outcomes to the numbers in the 1 st row of the triangle: One Coin tossed Number of heads Number of equivalent outcomes 1 st possible outcome: H nd possible outcome: T 0 1 Uses for Pascal s Triangle, page 2

3 Suppose you toss 2 coins. Now cover the numbers in the 2 nd row to represent the number equivalent outcomes when tossing 2 coins. The following are the possible outcomes: Look at the number of heads in these possible outcomes: 1 st coin 2 nd coin 1 st possible outcome: H H 2 nd possible outcome: H T 3 rd possible outcome: T H 4 th possible outcome: T T Compare the number equivalent outcomes to the numbers in the 2 nd row of the triangle: Number of heads Number of equivalent outcomes The probability of getting 2 heads from tossing 2 coins is: 1 chance out of 4 outcomes = ¼. The probability of getting one head and one tail from tossing 2 coins is 2 chances out of 4 = ½. The probability of getting zero heads (= 2 tails) is one chance out of 4 = ¼. Taking this further, toss 3 coins, and cover up the 3 rd row of numbers. The following tree diagram represents the possible outcomes: The probability of getting exactly 3 heads after tossing 3 coins is 1 chance out of 8, or 1/8. The probability of getting exactly 2 heads is 3 chances out of 8, or 3/8. Uses for Pascal s Triangle, page 3

4 Look at the number of heads in the possible outcomes: 1 st coin 2 nd coin 3 rd coin 1 st possible outcome: H H H 2 nd possible outcome: H H T 3 rd possible outcome: H T H 4 th possible outcome: H T T 5 th possible outcome: T H H 6 th possible outcome: T H T 7 th possible outcome: T T H 8 th possible outcome: T T T Compare now the number of heads to the number of equivalent outcomes: Number of heads Number of equivalent outcomes By now you might recognize that the number of equivalent outcomes are the same as the numbers in the third row of Pascal s Triangle!!! If we continue by tossing n coins we would end up with the number of equivalent outcomes of heads being the same as the numbers in the n th row of Pascal s Triangle! For example, if 4 coins are tossed, then the numbers of equivalent outcomes of heads are the numbers which are also the numbers in the 4 th row of Pascal s Triangle! Predict your odds of getting heads with coin tossing patterns in Pascal s Triangle!!! The Fibonnacci Sequence! Fibonnacci's Sequence of numbers starts with 1, 1, and then the next number is added to the previous number to become the next number in the sequence and so on. This creates: 1,1,2,3,5,8,13,21,34, 55,89,144,233, etc.... This sequence can also be located in Pascal's Triangle! Put a token on the first 1 (in row 0) on the triangle. This represents marking off that number as the first number in the Fibonnaci sequence. Next, use tokens to cover up numbers (see illustration below) that appear along each straight line (perhaps use a different colored playing piece on each straight line). The numbers in the consecutive rows shown in the diagram below are the first numbers of the Fibonnacci Sequence. The Fibonnacci Sequence can be found in many places, including the Golden Ratio, the lengths of the segments of a pentagram, the forms of plants and shells, harmonies played on musical instruments, and more! Uses for Pascal s Triangle, page 4

5 Sierpinski's Gasket Put a token on all the odd numbers in Pascal's Triangle, and leave the rest alone. What pattern do you see? You will find the recursive Sierpinski Gasketfractal is revealed (see figure below), yet another fascinating pattern in Pascal's Triangle! Other interesting patterns are formed if the elements not divisible by other numbers are filled, especially those indivisible by prime numbers. For example, cover only the numbers in Pascal s triangle that are multiples of 3, and then do this again after clearing the board for multiples of 5, and of 7. What do you see? See the related RAFT Idea Sheets: Sierpinski Gasket - Gasket.pdf Tetrix - Square Rings Place a token on each of six numbers surrounding a number on Pascal s Triangle. What do you notice? For example (assume shaded squares are covered by tokens): 1 x 1 x 5 x 10 x 6 x 3 = 900 = x 4 x 10 x 10 x 4 x 3 = 14,400 = The product of the surrounding number is a perfect square! Also the products of alternate triples are equal! e.g., 1 x 5 x 6 = 1 x 10 x 3 = 30 and 3 x 10 x 4 = 4 x 10 x 3 = 120. Can you find a relationship between the number that is surrounded and the value of the squared number? Uses for Pascal s Triangle, page 5

6 Figures Formed by Points on a Circle Study the numbers in the chart below. What do you notice???? The numbers are actually the top of Pascal's Triangle, except the first 1 (element zero) in each row is missing. The figures are formed by placing a number of points on the circumference of a circle and then drawing all the possible lines between them(it does not matter if the points are equally spaced along the circumference). For instance, for a figure with n points, look at the n th row of Pascal s Triangle in order to find the number of points, line segments, and polygons of the figure formed by having its vertices on the circle! So, if a pentagon is formed by 5 points (vertices) on the circle, then the 5 th row of Pascal s Triangle shows the number of points (which is 5), the number of line segments (which are 10), the number of triangles formed within the figure by line segments (which is another 10), the number of quadrilaterals formed within the figure by line segments (which are 5), and the number of pentagons formed (which is one). There are no 6-sided hexagons or 7-sided heptagons, or any n-sided figures for n < 5. Uses for Pascal s Triangle, page 6

7 Binomial Expansions! Consider the following expanded powers of (a + b) n, where a + b is any binomial and n is a whole number. Look for patterns! Unexpanded Form Expanded Form (a + b) 0 = 1 (a + b) 1 = a + b (a + b) 2 = a 2 + 2ab + b 2 (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 (a + b) 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 (a + b) 5 = a 5 + 5a 4 b + 10a 3 b a 2 b 3 + 5ab 4 + b 5 (note: this pattern could go on infinitely!) Each expanded form is a polynomial. Do you see any relationships to Pascal s Triangle? 1. Notice Pascal s Triangle is embedded within these polynomial expansions! It turns out knowing Pascal s Triangle can help you write the expanded form of a polynomial without much effort! For example, how can you expand the polynomial (x + 2) 3? Well, you could use the technique of multiplying the factors together, which can take time. But according to Pascal's triangle, in the expanded polynomial for (a + b) 3 the coefficients of each term are 1(from 1a ), 3 (from 3a b), 3 (from 3ab ), and 1 (from b 3 ). This means that for (x + 2) 3, x is a, and 2 is b, so by plugging in these values you get: (x + 2) 3 = x 3 + (3x 2 )(2) + (3x)(2 2 ) = x 3 + 6x x + 8, the expanded form! 2. When n is even, ( b) n will be positive. But when n is odd, ( b) n will be negative. This means that each odd power of b will have a negative sign. Look at the following example for using Pascal s triangle to expand (x 1) 5 Again, looking at Pascal's triangle, the binomial coefficients for (a + b) 5 are In this case, x is a, and 1 is "b". The signs will alternate, and each odd power of 1 will have a negative sign : (x 1) 5 = x 5 (-1) 0 + 5x 4 (-1) x 3 (-1) x 2 (-1) 3 + 5x (-1) 4 + x 0 (-1) 5 = x 5 5x x 3 10x 2 + 5x 1, the expanded form! Uses for Pascal s Triangle, page 7

8 What other ways does Pascal s Triangle relate to expanded polynomials? 3. For each term in an expanded polynomial, the sum of the exponents is n, the power to which the binomial is raised. For example, in (a + b) 4, add the exponents in each term: for a 4, which is a 4 b 0, add the exponents to get 4; for each 4a 3 b add the exponents to get 4; for each 6a 2 b 2 add the exponents to get 4; and for b 4, which is a 0 b 4, add the exponents to get Also, notice for every polynomial expansion, there is one more term than the power of the exponent, n. For example, (a + b) 2 has 3 terms in its expanded form: a 2 + 2ab + b 2, which are 1, 2, and 1, and the power of the exponent is 2. For (a + b) 5, there are 6 terms (1, 5, 10, 10, 5, and 1); again one more term than the power of the exponent In each expanded polynomial, the exponents of the first term a begin with n, the power of the binomial, and the power of each successive a in each term decreases by one to the power 0 in the last term. The last term has no factor of a (since a 0 = 1). Likewise, first term has no factor of b (since b 0 = 1), and the powers of b start with 0 and increase term by term to the power of n. 6. The coefficients for the terms start at 1 and increase through certain values about "half"-way and then decrease through these same values back to 1. Uses for Pascal s Triangle, page 8