Exam s 3-4 Solutions

Exam s 3-4 Solutions - MATH 290 EXAM 3 SOLUTIONS n 1(16 pts...

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MATH 290 – EXAM 3 – SOLUTIONS 1. (16 pts.) Prove by induction that for all natural numbers n , n X k =1 ( k · k !) = ( n + 1)! - 1. Solution: Assume that n = 1. Then 1 X k =1 ( k · k !) = 1 · 1! = 1 = 2! - 1. Hence the result holds for n = 1. Let n be any natural number and assume that n X k =1 ( k · k !) = ( n + 1)! - 1. We must show that n +1 X k =1 ( k · k !) = ( n + 2)! - 1. By the induction hypothesis, n +1 X k =1 ( k · k !) = n X k =1 ( k · k !) + ( n + 1)( n + 1)! = ( n + 1)! - 1 + ( n + 1)( n + 1)! . Also, ( n +1)! - 1+( n +1)( n +1)! = ( n +1)![1+( n +1)] - 1 = ( n +1)!( n +2) - 1 = ( n +2)! - 1 . This is what we wanted to prove. Therefore by the Principle of Mathematical Induction, for all natural numbers n , n X k =1 ( k · k !) = ( n + 1)! - 1. 2. (16 pts.) Prove by induction that for all natural numbers n 4, 2 n < n !. Solution: Assume that n = 4. Then 2 n = 2 4 = 16 < 24 = 4!. Hence the result holds for n = 4. Let
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Exam s 3-4 Solutions - MATH 290 EXAM 3 SOLUTIONS n 1(16 pts...

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