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Mathematics, 15.04.2020 03:28 oshawn108

Prove the following statement using mathematical induction. Do not derive it from Theorem 5.2.1 or Theorem 5.2.2. For every integer n β‰₯ 1, 1 + 6 + 11 + 16 + + (5n βˆ’ 4) = n(5n βˆ’ 3) 2 . Proof (by mathematical induction): Let P(n) be the equation 1 + 6 + 11 + 16 + + (5n βˆ’ 4) = n(5n βˆ’ 3) 2 . We will show that P(n) is true for every integer n β‰₯ 1. Show that P(1) is true: Select P(1) from the choices below. P(1) = 1 1 = 1 Β· (5 Β· 1 βˆ’ 3) 2 1 + (5 Β· 1 βˆ’ 4) = 1 Β· (5 Β· 1 βˆ’ 3) P(1) = 1 Β· (5 Β· 1 βˆ’ 3) 2 The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer k β‰₯ 1, if P(k) is true, then P(k + 1) is true: Let k be any integer with k β‰₯ 1, and suppose that P(k) is true. The left-hand side of P(k) is and the right-hand side of P(k) is . [The inductive hypothesis states that the two sides of P(k) are equal.] We must show that P(k + 1) is true. P(k + 1) is the equation 1 + 6 + 11 + 16 + β‹― + (5(k + 1) βˆ’ 4) = . After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes + (5(k + 1) βˆ’ 4). When the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal . Hence P(k + 1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]

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