Page 374 - HOW TO PROVE IT: A Structured Approach, Second Edition

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360 Appendix 1: Solutions to Selected Exercises
(Notice that the second to last step uses both the inductive hypothesis
and part (a).)
n
0
15. Base case: n = 0. Then a n = a 0 = 0 = 2 − 0 − 1 = 2 − n − 1.
n
Induction step: Suppose that n ∈ N and a n = 2 − n − 1. Then
n
a n+1 = 2a n + n = 2(2 − n − 1) + n
= 2 n+1 − 2n − 2 + n = 2 n+1 − n − 2 = 2 n+1 − (n + 1) − 1.
n
n
n! n!
18. (a) = = 1 and = = 1.
0 0!·n! n n!·0!
(b)

n n n! n!
+ = +
k k − 1 k!(n − k)! (k − 1)!(n − k + 1)!
n!(n − k + 1) n!k
= +
k!(n − k + 1)! k!(n − k + 1)!
n!(n + 1) n + 1
= = .
k!(n + 1 − k)! k
(c) We follow the hint.
Base case: n = 0. Suppose A is a set with 0 elements. Then A = ∅,
the only value of k we have to worry about is k = 0, P 0 (A) ={∅},
0

which has 1 element, and = 1.
0
Induction step: Suppose the desired conclusion holds for sets with
n elements, and A is a set with n + 1 elements. Let a be an element of
A, and let A = A \{a}, which is a set with n elements. Now suppose

0 ≤ k ≤ n + 1. We consider three cases.
Case 1: k = 0. Then P k (A) ={∅}, which has 1 element, and
n+1 = 1.

k
Case 2: k = n + 1. Then P k (A) ={A}, which has 1 element, and
n+1 = 1.

k
Case 3. 0 < k ≤ n. There are two kinds of k-element subsets of
A: those that contain a as an element, and those that don’t. The k-
element subsets that don’t contain a are just the k-element subsets
n

of A , and by inductive hypothesis there are of these. Those that

k

do contain a are the sets of the form X ∪{a}, where X ∈ P k−1 (A ),
n
and by inductive hypothesis there are of these, since this is the
k−1
number of possibilities for X. Therefore by part (b), the total number
of k-element subsets of A is

n n n + 1
+ = .
k k − 1 k