## Question

###### Prove the following statement using mathematical induction. Donot derive it from Theorem 5.2.1 or Theorem 5.2.2.For every integer n â‰¥ 1, 1 + 6 + 11 + 16 + + (5n âˆ’ 4) = n(5n âˆ’ 3)2.Proof (by mathematical induction):Let P(n) be the equation1 + 6 + 11 + 16 + + (5n âˆ’ 4) = n(5n âˆ’ 3)2.We will show that P(n) is true for everyinteger n â‰¥ 1.Show that P(1) is true: Select P(1) from the choices below.1 + (5 Â· 1 âˆ’ 4) =1 Â· (5 Â· 1 âˆ’ 3)1 = 1 Â· (5 Â· 1 âˆ’ 3)2 P(1) = 5 Â· 1 âˆ’ 4P(1)

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 â‰¥ 1, 1 + 6 + 11 + 16 + + (5n âˆ’ 4) = n(5n âˆ’ 3) 2 . Proof (by mathematical induction): Let P(n) be the equation 1 + 6 + 11 + 16 + + (5n âˆ’ 4) = n(5n âˆ’ 3) 2 . We will show that P(n) is true for every integer n â‰¥ 1. Show that P(1) is true : Select P(1) from the choices below. 1 + (5 Â· 1 âˆ’ 4) = 1 Â· (5 Â· 1 âˆ’ 3) 1 = 1 Â· (5 Â· 1 âˆ’ 3) 2 P(1) = 5 Â· 1 âˆ’ 4 P(1) = 1 Â· (5 Â· 1 âˆ’ 3) 2 The selected statement is true because both sides of the equation equal . Show that for each integer k â‰¥ 1, if P(k) is true, then P(k + 1) is true : Let k be any integer with k â‰¥ 1, and suppose that P(k) is true. The left-hand side of P(k) is ---Select--- 5k âˆ’ 4 1 + (5k âˆ’ 4) 1 + 6 + 11 + 16 + â‹¯ + (5k âˆ’ 4) , 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 + â‹¯ + (5(k + 1) âˆ’ 4) = . After substitution from the inductive hypothesis, the left-hand side of P(k + 1) becomes ---Select--- k(5k âˆ’ 3)/2 ((k âˆ’ 1)(5k âˆ’ 3))/2 ((k + 1)(5k âˆ’ 3))/2 ((k âˆ’ 1)(5(k âˆ’ 1) âˆ’ 3))/2 + (5(k + 1) âˆ’ 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.]