Use the Principle of Mathematical Induction to show that the following statement is true for all natural numbers n. 6 + 12+18 + ... + 6n = 3n(n +1) What two conditions must the given statement satisfy to prove that it is true for all natural numbers? Select all that apply. The statement is true for the natural number 1. 0 The statement is true for any two natural numbers k and k + 1. 0 If the statement is true for the natural number 1, it is also true for the next natural number 2. If the statement is true for some natural number k, it is also true for the next natural number k + 1. Show that the first of these conditions is satisfied by evaluating the left and right sides of the given statement for the first natural number. 6+ 12 +18 + ...+ 6n = 3n(n +1) =(Simplify your answers.) To show that the second condition is satisfied, write the given statement for k + 1. 6+12 + ...+6k += (Simplify your answers. Type your answers in factored form.) If the statement for k + 1 is true whenever the given statement 6 + 12 + 18+ ...+ 6n = 3n(n + 1) is true for all natural numbers. Use the statement for k, 6 + 12 + ... + 6k = 3k(k + 1), to simplify the left side. K(K + 1) + (K +1)=3(K +1)(k+2) Use the distributive rule and the associative rule to rewrite the right side. 3k(k + 1) +6(k + 1) = k(k+1) +(k+1) I Ise this result to draw a ronchisinn regarding the niven statement 6+ 12 + 18 + ... + fin = 3nin+1) Time Remaining: 02:22:36