Stitz-Zeager_College_Algebra_e-book

# Since 270 represents 3 of a 4 counter clockwise

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Unformatted text preview: = (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 n(n + 1)(2n + 1) is true for all natural numbers n ≥ 1. 6 (d) Let P (n) be the sentence 3n > n3 . Our base case is n = 4 and we check 34 = 81 and 43 = 64 so that 34 > 43 as required. We now assume P (k ) is true, that is 3k > k 3 , and try to show P (k + 1) is true. We note that 3k+1 = 3 · 3k > 3k 3 and so we are done if we can show 3k 3 > (k + 1)3 for k ≥ 4. We can solve the inequality 3x3 > (x + 1)3 using the techniques of Section 5.3, and doing so gives us x > √ 1 ≈ 2.26. Hence, for k ≥ 4, 3 3−1 3k+1 = 3 · 3k > 3k 3 > (k + 1)3 so that 3k+1 > (k + 1)3 . By induction, 3n > n3 is true for all natural numbers n ≥ 4. 580 Sequences and the Binomial Theorem (f) Let P (n) be the sentence log (xn ) = n log(x). For the duration of this argument, we assume x > 0. The base case P (1) amounts checking that log x1 = 1 log(x) wh...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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