Explanation Given information: assumen is an integer variable that takes positive integer values. Proof: 2 n 3 + 2 n 3 2 + 2 n 3 3 + ... + 2 n 3 n = 2 n ⋅ ( 1 3 + 1 3 2 + 1 3 3 + ... + 1 3 n ) Distributive property = 2 n ⋅ ( 1 + 1 3 + ( 1 3 ) 2 + ... + ( 1 3 ) n − 1 ) = 2 n ⋅ ( ∑ k = 0 n ( 1 3 ) k − 1 ) = 2 n ⋅ ( ∑ k = 0 ( n + 1 ) − 1 ( 1 3 ) k − 1 ) = 2 n ⋅ ( ( 1 3 ) n + 1 − 1 1 3 − 1 − 1 ) ∑ k = 0 n − 1 a k = a n − 1 a − 1 = 2 n ⋅ ( ( 1 3 ) n + 1 − 1 − 2 3 − 1 ) = n ⋅ ( 3 − ( 1 3 ) n − 1 ) = n ⋅ ( 2 − ( 1 3 ) n ) Since − ( 1 3 ) n ≤ 0 : | n ⋅ ( 2 − ( 1 3 ) n ) <

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Chapter 11.4, Problem 37ES

To determine

To prove:

2n3+2n32+2n33+...+2n3n is Θ(n).

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Chapter 11.4, Problem 36ES

Chapter 11.4, Problem 38ES

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Chapter 11 Solutions

Discrete Mathematics With Applications

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