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Chapter 1 Automata CE year IV.

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Presentation on theme: "Chapter 1 Automata CE year IV."— Presentation transcript:

1 Chapter 1 Automata CE year IV

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16 Mathematical induction can be used to prove that the following statement, which we will call P ( n ), holds for all natural numbers n. P ( n ) gives a formula for the sum of the natural numbers less than or equal to number n. The proof that P ( n ) is true for each natural number n procee ds as follows.natural numbers Basis : Show that the statement holds for n = 0. P (0) amounts to the statement: In the left-hand side of the equation, the only term is 0, and so the left-ha nd side is simply equal to 0. In the right-hand side of the equation, 0·(0 + 1)/2 = 0. The two sides are equal, so the statement is true for n = 0. Thus it has be en shown that P (0) holds.

17 Inductive step: Show that if P(k) holds, then also P(k + 1) hol ds. This can be done as follows. Assume P(k) holds (for some unspecified value of k). It must then be shown that P(k + 1) holds, that is: Using the induction hypothesis that P(k) holds, the left-hand s ide can be rewritten to: Algebraically: thereby showing that indeed P(k + 1) holds.

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19 Example: Using mathematical induction to prove that 1 + 5 + 9 + ----- - + (4n - 3) = n(2n - 1) for all positive integers. Solution: Step 1: Let P n be the statement 1 + 5 + 9 + ---- + (4n - 3) = n (2n - 1). Step 2: P1 is true because (4(1) - 3) = 1(2(1) - 1) [The Anchor.] Step 3: Assume that Pk is true, so that Pk: 1 + 5 + 9 + ---- + (4k - 3) = k (2k - 1) [The Inductive Hypothesis.] Step 4: The next term on the left-hand side would be (4(k + 1) - 3). Step 5: 1 + 5 + 9 + ---- + (4k - 3) + (4(k + 1) - 3) = k (2k - 1) + (4(k + 1) - 3) Step 6: = 2k2 - k + 4k + 1 Step 7: = 2k2 + 3k + 1 Step 8: = (k + 1)2 + k (k + 1) Step 9: = (k + 1) (2k + 1) [Factor.] Step 10: = (k + 1) (2(k + 1) - 1) Step 11: = P k +1 Step 12: So, the equation is true for n = k + 1. Step 13: Therefore, Pk is true for all positive integers, by mathematic al induction.

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35 1. Assignment 1. prove the following: 1.1/1*2+1/2*3+1/3*4+……1/n(n+1)=n/n+1 all n>=1 2.1+4+7+..+(3n-2)=n(3n-1)/2 for all n>=1 3.n 3 +2n is divisible by 3 for all positive values of n 4.2+2 2 +2 3 +..+2 n =2(2n-1) for all n 5.2 n >=1+n for all n 6.5 n -2 n is divisible by 3 for all positive integer value of n Deadline :7 days from the date of instruction: End of Chapter 1


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