Click and drag the steps in the correct order to show that 3 divides 3 + 2n whenever n is a positive integer using mathematical induction. BASIS STEP: By the inductive hypothesis, 31 (K + 2k), and certainly 3 131K2 + 1). (k+ 1) + 2(k+ 1) = (k+ 3k + 1) + (2k + 2) = (k® + 2k) + 3(K + 1) INDUCTIVE STEP: Suppose that 31 (K® + 2K). As the sum of two multiples of 3 is again divisible by 3, 31 ((k+ 1) + 2(k + 1)). 31 (13 + 2 · 1), i.e., 313, so the basis step is true. (k+ 1)3 + 2(k+ 1) = (ké + 3k2 + 3k + 1) + (2k + 2) = (k® + 2K) + 3(k2 + k + 1) 31 (03 + 2.0), i.e., 310, so the basis step is true. By the inductive hypothesis, 31 (K + 2k), and certainly 313(K + k + 1). Reset