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Unformatted text preview: Discrete Mathematics Dr. Fred Phelps Lecture 6 Pascals Triangle and the Normal LPV 3.5 Pascals Triangle n th row consists , ,... , . 1 1 n n n n n n  LPV 3.6 Pascals Triangle  identities Recall the identities we proved earlier: 1) The triangle is symmetric about the center, i.e. 2) The sum across the n th row is 2 n , i.e. 2 . n n i n i = = . n n k n k =  1 1 . 1 n n n k k k = +  3) Every number is the sum of the two numbers above it in the triangle, i.e. LPV 3.6 Pascals Triangle  identities Consider the sum with signs alternating 1 1 with 1 n n n k k k +  ... ( 1) . 1 2 3 n n n n n n n ++ +  Replace for k =1, 2, . . n1 in this sum. Also replace 1 1 with and with 1 n n n n n n  LPV 3.6 Pascals Triangle  identities ... ( 1) 1 2 3 n n n n n n n ++ + = 1 1 1 1 1 1 1 1 1 2 2 3 n n n n n n n  + + + 1 1 1 1 ... ( 1) ( 1) 2 1 1 n n n n n n n n  + + + +   Then The entire sum is zero because the second term cancels the first, the fourth cancels the third, and so on down to the very last pair of 0. = LPV 3.6 Pascals Triangle  identities ... ( 1) 1 2 3 n n n n n n k ++ + = 1 1 1 1 1 1 1 1 1 2 2 3 n n n n n n n  + + ++ 1 1 ... ( 1) 1 k n n k k  + + +  If we stop this sum at the k +1st term All the terms except the last one cancel giving 1 ( 1) ....
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This note was uploaded on 10/15/2011 for the course MATHS 100 taught by Professor Fredphelps during the Spring '11 term at Jordan University of Science & Tech.
 Spring '11
 FredPhelps
 Math

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