In discovering the expression 2 for the number of subsets of a set with n elements, it was observed that for the first few values of n. increasing the number of elements by one doubles the number of subsets. Here, prove the formula in general by showing that the same is true for any value of n. Assume set A has n elements and s subsets. Now add one additional element, say e, to the set A. (Now there is a new set, say B. with n + 1 elements.) Divide the subsets of B into those that do not contain e and those that do. Complete parts (a) through (d). Here is a summary of what has been done. (1) It was observed for the first few values of n (say n = 1, 2, 3) that a set with n elements has 2 n subsets. It was shown that, in general, whenever that observation holds for a particular value of n, it also must hold for the (2) next value, n +1. (3) Therefore, what was observed for n = 1. 2, and 3 also must hold for n =4, hence for n = 5, and so on without limit. Therefore, the result must hold for all counting numbers n. This process of proof is called mathematical induction. (a) How many subsets of B do not contain e? (Hint Each of these is a subset of the original set A.) There are subsets of B that do not contain (Type an expssion using s as the variable. Simplify your answer