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# sole5 - ANALYSIS EXERCISE 5SOLUTIONS 1(a Base case n = 10...

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ANALYSIS EXERCISE 5–SOLUTIONS 1. (a) Base case : n = 10; 2 10 = 1024 > 1000 = 10 3 . The statement is true. Induction step: Suppose that 2 k > k 3 ( k 10). We have to prove that 2 k +1 > ( k + 1) 3 . By the assumption 2 k +1 = 2 · 2 k > 2 k 3 . So it is sufficient to prove that 2 k 3 > ( k +1) 3 which is equivalent to k 3 - 3 k 2 - 3 k - 1 > 0. Since k 10 we have that k - 3 7 and k 2 100. Therefore we have k 3 - 3 k 2 - 3 k - 1 = ( k - 3)( k 2 - 3) - 10 which is evidently positive. This proves the statement for n = k +1. By the principle of mathematical induction the statement is true for all natural n 10. (b) Base case : n = 9; 2 9 = 512 > 4 · 9 2 + 1. The statement is true. Induction step: Suppose that 2 k > 4 k 2 + 1 ( k 9). We have to prove that 2 k +1 > 4( k + 1) 2 + 1. By the assumption 2 k +1 = 2 · 2 k > 2(4 k 2 + 1), so it is sufficient to prove that 2(4 k 2 + 1) > 4( k + 1) 2 + 1 which is easy to see taking the difference between the left-hand side and the right-hand side. This proves the statement for n = k + 1. By the principle of mathematical induction the statement is true for all natural n 9.

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