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) 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) 0=(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. 3 k(k + 1)+6(k +1) = K(k+1) +(k+1) Use this result to draw a conclusion regarding the given statement, 6 + 12 + 18+ ... + 6n = 3n(n +1). A. Since this statement cannot be shown to be true for all values of k, the second condition required to prove that the given statement is true for all natural numbers is not satisfied. However, since at least one condition in the Principle of Mathematical Induction is satisfied, the given statement is true for all natural numbers. O B. Since the right side of the statement for k + 1 simplifies to the left side of the statement for k, the second condition required to prove that the given statement is true for all natural numbers is satisfied, and the given statement is true for all natural numbers. O C. Since this statement is true for all values of k, the second condition required to prove that the given statement is true for all natural numbers is satisfied. The first condition in the Principle of Mathematical Induction is also satisfied, so the given statement is true for all natural numbers