Click and drag the steps in the correct order to show that 3 divides m" + 2n whenever n is a positive integer using mathematical induction. BASIS STEP: By the inductive hypothesis, 3 | (4* + 2k) , and certainly 3 | 3(47 + k + 1). (k + 1)3 + 2(k+ 1) = (k3 + 3/2 + 3/ + 1) + (2k + 2) = (k + 2k) + 3(k= +k + 1)0 INDUCTIVE STEP: (k + 1)3 + 2(k + 1) = (43 + 3/2 + 1) + (2k + 2) = (43 + 2k) + 3(kz + 1) As the sum of two multiples of 3 is again divisible by 3, 3 | ((k + 1)3 + 2(k + 1)). By the inductive hypothesis, 3 | (13 + 2k) , and certainly 3 | 3(12 + 1). Suppose that 3 | (13 + 2k). 3 | (03 + 2 . 0), i.e., 3 | 0, so the basis step is true. 3 | (13 + 2 . 1), i.e., 3 | 3, so the basis step is true