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# key5sp00 - COT3100-01 Spring 2000 S Lang Solution Key to...

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COT3100-01, Spring 2000 Assigned: 4/12/2000 S. Lang Solution Key to Assignment #5 (40 pts.) Due: 4/19 in class 1. Use induction on n 0 to prove the following summation identity: . 0 2 ) 2 ( ) 1 ( 2 ) 1 ( = + - = - n i n n n i i We use induction on n 0. (Basis Step) Consider n = 0. In this case, . 0 2 0 2 ) 0 2 0 ( 0 ) 1 ( RHS The . 0 0 ) 1 ( ) 1 ( LHS The 2 0 0 0 2 = = + - = = - = ∑ - = = i i i Thus, LHS = RHS, so the Basis Step is proved. (Induction Hypothesis) Consider n = k . Suppose . 0 some for 0 2 ) 2 ( ) 1 ( 2 ) 1 ( = + - = - k k i k k k i i (Induction Step) Consider n = k + 1. We need to prove (1). of RHS 2 ) 2 3 ( 1 ) 1 ( 2 ) 2 4 2 2 ( 1 ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( since , 2 ) 1 ( 2 ) 1 ( 2 ) 2 )( 1 ( 1 ) 1 ( Hypothesis Induction by the , ) 1 ( ) 1 ( 2 ) 2 ( ) 1 ( summation of definition by the , ) 1 ( ) 1 ( ) 1 ( (1) of LHS The 2 2 1 2 2 1 2 1 2 1 0 2 = + + + - = + + + - - + - = - - = - = - + - + + - + - = + - + + - = + - + ∑ - = + + + + + = k k k k k k k k k k k k k k k k k i k k k k k k k i i Thus, the Induction Step is proved. By induction, we proved that 2. Suppose a function f ( n ) is defined for integer n 0 recursively by the following recurrence: f (0) = 7; f (1) = 14; and f ( n ) = f ( n – 1) + 12 f ( n – 2) for n 2. Prove that f ( n ) = 5(4) n + 2(–3) n for all n 0. We use strong induction on n 0. (Basis Step) We first consider n = 0. In this case, applying the formula for f ( n ) yields f (0) = 5(4) 0 + 2(–3) 0 = 5 + 2 = 7, which equals the given initial value for f (0).

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key5sp00 - COT3100-01 Spring 2000 S Lang Solution Key to...

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