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 2 1,1+6+11+16++ (Sn-4)= n(n-3) Proof (by mathematical induction): Let P(n) be the equation 1 +6+11 +16+ +(5n-4) = n(5n-3) 2 We will show that P(n) is true for every integer n 2 1. Show that P(1) is true: Select P(1) from the choices below. 01+(5-1-4) 1 (5-1- 3) 1-1-(5-1-3) 2 O P(1) = 1·(5 - 1-(5-1-3) 2 OP(1)-5-1-4 The selected statement is true because both sides of the equation equal 1 Show that for each integer k 2 1, if P(k) is true, then P(k+ 1) is true: Let k be any integer with k 2 1, and suppose that P(K) is true. The left-hand side of P(K) is 1 + 6 + 11 + 16 + (Sk-4) X, and the right-hand side of P(k) is 1 [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 + + (5(k+1)-4)= +6+11+16++ right-hand sides of P(k+ 1) are simplified, they both can be shown to equal 1 (k+ 1) (5(k+1)-3) 2 Hence P(k+ 1) is true, which completes the inductive step. X [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] x After substitution from the inductive hypothesis, the left-hand side of P(x + 1) becomes (k(5k - 3)/2 ✔+(5(k+1)-4). When the left-hand and