# hw9 - CSE 260 Homework 9 Induction ANSWER 1 Section 4.3 20...

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CSE 260 Homework 9- Induction- ANSWER October 22, 2007 1. Section 4.3: 20 max ( a 1 , a 2 , ...a n +1 ) = max ( max ( a 1 , a 2 , ...a n ) , a n +1 ) min ( a 1 , a 2 , ...a n +1 ) = min ( min ( a 1 , a 2 , ...a n ) , a n +1 ) 2. Prove that 5 n - 4 n - 1 is divisible by 16 for n²Z + p ( n ) : 5 n - 4 n - 1 is divisible by 16 for n²Z + p (1) : 5 1 - 4 . 1 - 1 is divisible by 16 ? 0 is divisible by 16? true. p ( n + 1) : 5 n +1 - 4( n + 1) - 1 is divisible by 16? 5 n +1 - 4( n + 1) - 1 = 5(5 n - 4 n - 1) + 16 n is divisible by 16? yes. 3. Recursively deFne b 0 = b 1 = b 2 = 1 and b n = b n - 1 + b n - 3 for n 3. (a) Calculate b n for n = 3 , 4 , 5 b 3 = b 2 + b 0 = 1 + 1 = 2 b 4 = b 3 + b 1 = 2 + 1 = 3 b 5 = b + 4 + b 2 = 3 + 1 = 4 (b) Show by using the second principle (strong) of mathematical induc- tion that b n 2 b n - 2 for n 3 proof: b n +1 2 b n - 1 for n + 1 3 b n +1 = b n + b n - 2 2 b n - 2 + 2 b n - 4 = 2( b n - 2 + b n - 4 ) = 2 b n - 1 4. Prove by mathematical induction 1 3 + 2

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hw9 - CSE 260 Homework 9 Induction ANSWER 1 Section 4.3 20...

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