Pascals Triangle MCR3U: Functions Pascals Triangle is an - PDF document

s e q u e n c e s a n d s e r i e s s e q u e n c e s a n d s e r i e s Pascal’s Triangle MCR3U: Functions Pascal’s Triangle is an arrangement of numbers, generated using a simple iterative process. While Pascal’s Triangle was not “invented” by Blaise Pascal, he is credited for applying it toward probability theory. Begin with a triangular arrangement of 1s, as shown. Pascal’s Triangle and the Binomial Theorem 1 J. Garvin 1 1 Continue to place 1s at the outer edges of each new row. Each interior value is the sum of the two values diagonally above it. J. Garvin — Pascal’s Triangle and the Binomial Theorem Slide 1/18 Slide 2/18 s e q u e n c e s a n d s e r i e s s e q u e n c e s a n d s e r i e s Pascal’s Triangle Pascal’s Triangle 1 The first row in Pascal’s Triangle is Row 0. This means that the n th row has n + 1 entries. 1 1 Each column is read diagonally downwards, from right to left. 1 2 1 The first column is Column 0. 1 3 3 1 A entry’s position may be denoted t n , r , indicating the r th 1 4 6 4 1 term in the n th row. 1 5 10 10 5 1 For example, t 4 , 1 = 4 and t 5 , 3 = 10, as shown on the previous slide. etc. J. Garvin — Pascal’s Triangle and the Binomial Theorem J. Garvin — Pascal’s Triangle and the Binomial Theorem Slide 3/18 Slide 4/18 s e q u e n c e s a n d s e r i e s s e q u e n c e s a n d s e r i e s Pascal’s Triangle Pascal’s Triangle t 0 , 0 Example t 1 , 0 t 1 , 1 Express t 4 , 1 + t 4 , 2 as a single term in Pascal’s Triangle. t 2 , 0 t 2 , 1 t 2 , 2 The term must be in the next row (Row 5) and the larger of t 3 , 0 t 3 , 1 t 3 , 2 t 3 , 3 the two columns (Column 2), so the term is t 5 , 2 . t 4 , 0 t 4 , 1 t 4 , 2 t 4 , 3 t 4 , 4 Example t 5 , 0 t 5 , 1 t 5 , 2 t 5 , 3 t 5 , 4 t 5 , 5 Express t 16 , 7 as the sum of two terms in Pascal’s Triangle. etc. The two terms that add to t 16 , 7 must come from the row Terms in Pascal’s Triangle above (Row 15), and the larger of the two columns will have the same value as the given term (7), so the sum is For any term, t n , r , in Pascal’s Triangle, � 1 t 15 , 6 + t 15 , 7 . if r = 0 or r = n t n , r = t n − 1 , r − 1 + t n − 1 , r otherwise J. Garvin — Pascal’s Triangle and the Binomial Theorem J. Garvin — Pascal’s Triangle and the Binomial Theorem Slide 5/18 Slide 6/18

s e q u e n c e s a n d s e r i e s s e q u e n c e s a n d s e r i e s Pascal’s Triangle Pascal’s Triangle There are many interesting patterns in Pascal’s Triangle. One Another interesting pattern in Pascal’s Triangle is often of these involves the sum of the entries in any given row. called “hockey stick” pattern. In the first row, there is only a 1, so the sum is 1. Beginning at the first entry in any column, sum the numbers downward and left to some arbitrary point, then move down In the second row, 1 + 1 = 2. and right one entry. In the third row, 1 + 2 + 1 = 4. 1 In the fourth row, 1 + 3 + 3 + 1 = 8. 1 1 Sum of the Entries in a Row of Pascal’s Triangle 1 2 1 The sum of all of the entries in the n th row of Pascal’s Triangle is 2 n . 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 J. Garvin — Pascal’s Triangle and the Binomial Theorem J. Garvin — Pascal’s Triangle and the Binomial Theorem Slide 7/18 Slide 8/18 s e q u e n c e s a n d s e r i e s s e q u e n c e s a n d s e r i e s Pascal’s Triangle Binomial Theorem Expand and simplify ( x + y ) 2 . The sum of the values in column 1 from t 1 , 1 to t 4 , 1 is 1 + 2 + 3 + 4 = 10. This value is the value of t 5 , 2 . ( x + y ) 2 = ( x + y )( x + y ) “Hockey Stick” Pattern in Pascal’s Triangle = x 2 + xy + xy + y 2 In Pascal’s Triangle, the sum of the first k entries in a = 1 x 2 + 2 xy + 1 y 2 column is t n , r + t n +1 , r + t n +2 , r + . . . + t n + k , r = t n + k +1 , r +1 . There are many more patterns in Pascal’s Triangle, but the Hmmmm. . . main focus in this course is its application toward expanding binomials. J. Garvin — Pascal’s Triangle and the Binomial Theorem J. Garvin — Pascal’s Triangle and the Binomial Theorem Slide 9/18 Slide 10/18 s e q u e n c e s a n d s e r i e s s e q u e n c e s a n d s e r i e s Binomial Theorem Binomial Theorem Expand and simplify ( x + y ) 3 . Expand and simplify ( x + y ) 4 . ( x + y ) 4 = 1 x 4 + 4 x 3 y + 6 x 2 y 2 + 4 xy 3 + 1 y 4 ( x + y ) 3 = ( x + y )( x + y ) 2 Can you spot the pattern? = ( x + y )( x 2 + 2 xy + y 2 ) = x 3 + 2 x 2 y + xy 2 + x 2 y + 2 xy 2 + y 3 1 = 1 x 3 + 3 x 2 y + 3 xy 2 + 1 y 3 1 1 1 2 1 1 3 3 1 Looks familiar. . . 1 4 6 4 1 J. Garvin — Pascal’s Triangle and the Binomial Theorem J. Garvin — Pascal’s Triangle and the Binomial Theorem Slide 11/18 Slide 12/18

s e q u e n c e s a n d s e r i e s s e q u e n c e s a n d s e r i e s Binomial Theorem Binomial Theorem The coefficients of the simplified polynomial expression Example correspond to entries in Pascal’s Triangle. Expand ( x + 5) 3 . The coefficients in the expansion of ( x + y ) n are the entries Using Pascal’s Triangle is faster than using the Distributive in Row n . Property. The exponents follow a predictable pattern too. ( x + 5) 3 = t 3 , 0 x 3 · 5 0 + t 3 , 1 x 2 · 5 1 + t 3 , 2 x 1 · 5 2 + t 3 , 3 x 0 · 5 3 The exponent for x decreases by 1, while the exponent for y increases by 1. = 1 · x 3 + 3 · 5 x 2 + 3 · 25 x + 1 · 125 In each term, the sum of the exponents is n . = x 3 + 15 x 2 + 75 x + 125 Binomial Theorem ( x + y ) n = t n , 0 x n y 0 + t n , 1 x n − 1 y 1 + t n , 2 x n − 2 y 2 + . . . + t n , n x 0 y n J. Garvin — Pascal’s Triangle and the Binomial Theorem J. Garvin — Pascal’s Triangle and the Binomial Theorem Slide 13/18 Slide 14/18 s e q u e n c e s a n d s e r i e s s e q u e n c e s a n d s e r i e s Binomial Theorem Binomial Theorem Example Example Expand (3 a + 4 b ) 3 . Expand (2 a + 3 b ) 4 . Use the substitutions x = 3 a , y = 4 b and n = 3. Use the substitutions x = 2 a , y = 3 b and n = 4. (2 a + 3 b ) 4 = t 4 , 0 (2 a ) 4 + t 4 , 1 (2 a ) 3 (3 b ) + t 4 , 2 (2 a ) 2 (3 b ) 2 (3 a + 4 b ) 3 = t 3 , 0 (3 a ) 3 + t 3 , 1 (3 a ) 2 (4 b ) + t 3 , 2 (3 a )(4 b ) 2 + t 3 , 3 (4 b ) 3 + t 4 , 3 (2 a )(3 b ) 3 + t 4 , 4 (3 b ) 4 = 1(27 a 3 ) + 3(9 a 2 )(4 b ) + 3(3 a )(16 b 2 ) + 1(64 b 3 ) = 27 a 3 + 108 a 2 b + 144 ab 2 + 64 b 3 =1(16 a 4 ) + 4(8 a 3 )(3 b ) + 6(4 a 2 )(9 b 2 ) + 4(2 a )(27 b 3 ) + 1(81 b 4 ) =16 a 4 + 96 a 3 b + 216 a 2 b 2 + 216 ab 3 + 81 b 4 J. Garvin — Pascal’s Triangle and the Binomial Theorem J. Garvin — Pascal’s Triangle and the Binomial Theorem Slide 15/18 Slide 16/18 s e q u e n c e s a n d s e r i e s s e q u e n c e s a n d s e r i e s Binomial Theorem Questions? Example Expand (4 a − 2 b ) 4 . Use the substitutions x = 4 a , y = − 2 b and n = 4. Note that y is negative. What will happen? (4 a − 2 b ) 4 = t 4 , 0 (4 a ) 4 + t 4 , 1 (4 a ) 3 ( − 2 b ) + t 4 , 2 (4 a ) 2 ( − 2 b ) 2 + t 4 , 3 (4 a )( − 2 b ) 3 + t 4 , 4 ( − 2 b ) 4 =1(256 a 4 ) + 4(64 a 3 )( − 2 b ) + 6(16 a 2 )(4 b 2 ) + 4(4 a )( − 8 b 3 ) + 1(16 b 4 ) =256 a 4 − 512 a 3 b + 384 a 2 b 2 − 128 ab 3 + 16 b 4 If y is negative, the terms in the expansion alternate signs. J. Garvin — Pascal’s Triangle and the Binomial Theorem J. Garvin — Pascal’s Triangle and the Binomial Theorem Slide 17/18 Slide 18/18