Prove the following statement by mathematical induction. For every integer n ≥ 0, 7 - 1 is divisible by 6. Proof (by mathematical induction): Let P(n) be the following sentence. 7n1 is divisible by 6. We will show that P(n) is true for every integer n ≥ 0. Show that P(0) is true: Select P(0) from the choices below. (7⁰-1) 16 O 1 is a factor of 70 - 1 O 6 is a multiple of 7⁰ - 1 061 (7⁰1) The truth of the selected statement follows from the definition of divisibility and the fact that 70-1 = Show that for each integer k ≥ 0, if P(k) is true, then P(k+ 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(k) from the choices below. O 6 is a multiple of 7k - 1 O 6 is divisible by (7k - 1) (7k-1) is divisible by 6 O 1 is a factor of 7k - 1 [This is P(k), the inductive hypothesis.] We must show that P(k+ 1) is true. Select P(k+ 1) from the choices below. O 6 is a multiple of 7k + 1 - 1 O (7k+11) is divisible by 6 O 1 is a factor of 7k +1 -1. 6 is divisible by (7k+11) By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7k - 1 = 6r, and so 7k = 6r + 1. Now 7k+11=7k.7-1. When 6r + 1 is substituted for 7k in the above equation, and the right-hand side is simplified, the result can be expressed in terms of k and r as follows. 7k+11= 6. This quantity is an integer because k and r are integers. Select the final sentence from the choices below. O Hence, 1 is a factor of (7k+1-1), and so P(k+1) is false, which completes the inductive step. O Hence, (7k+11) is divisible by 6, and so P(k+1) is false, which completes the inductive step. Hence, (7k+11) is divisible by 6, and so P(k+1) is true, which completes the inductive step. O Hence, 1 is a factor of (7k+1-1), and so P(k+1) is true, which completes the inductive step. O Hence, 6 is divisible by (7k+1-1), and so P(k+1) is true, which completes the inductive step. O Hence, 6 is a multiple of (7k+1 -1), and so P(k+1) is false, which completes the inductive step.