Prove that for all natural numbers n, 1 + 10 + 100 + ......+107-1 = (107 - 1) . Select the steps necessary for a proof by induction. When n = 1, 1 + 10 + 100 + ......+1072-1 = 1, 2(10 -1) = 1. When n = 2, 1 + 10 + 100 + ...... +107-1 = 1+ 10 = 11; -(102 - 1) = -(10- 1)(10 + 1) = 1 + 10 = 11. When n = 3, 1 + 10 + 100 + ...... +1072-1 = 1+ 10 + 100 = 111; 2(103 - 1) = (10 - 1)(100 + 10 + 1) = 1 + 10 + 100 = 111, and so on. Assume 1 + 10 + 100 + ...... +1072 = (1on+1 - 1). Assume the formula holds when n = k. 0 1 + 10 + 100 + .... +10k-1 = 1(10k - 1) Show that 1 + 10 + 100 + ...... +10* = (10k+1 - 1). Since the result is true for n = 1 and the k + 1 case is true assuming the k th case is true the formula holds for all cases. Since the k + 1 case is true assuming the k th case is true the formula holds for all cases. When n = 1, 1 + 10 + 100 + ......+1072-1 = 1 and _(101 - 1) = 1