Page 302 - HOW TO PROVE IT: A Structured Approach, Second Edition

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288 Mathematical Induction
n

(This is called the binomial theorem, so the numbers are some-
k
times called binomial coefficients.)
n

Note: Parts (a) and (b) show that we can compute the numbers con-
k
veniently by using a triangular array as in Figure 1. This array is called
Pascal’s triangle, after the French mathematician Blaise Pascal (1623–
1662). Each row of the triangle corresponds to a particular value of n, and
n

it lists the values of for all k from 0 to n. Part (a) shows that the first
k
and last number in every row is 1. Part (b) shows that every other number
is the sum of the two numbers above it. For example, the lines in Figure
2
3
2

1 illustrate that = 3 is the sum of = 2 and = 1.
2 1 2
n = 0: 1
n = 1: 11
n = 2: 1 2 1
n = 3: 1 3 3 1
n = 4: 1 4 6 4 1
:
:
Figure 1: Pascal’s triangle
19. For the meaning of the notation used in this exercise, see exercise 18.
n
n n
(a) Prove that for all n ∈ N, = 2 . (Hint: You can do this by
k=0 k
induction using parts (a) and (b) of exercise 18, or you can combine
part (c) of exercise 18 with exercise 10 of Section 6.2, or you can
plug something in for x and y in part (d) of exercise 18.)

n k n
(b) Prove that for all n ≥ 1, (−1) = 0.
k=0 k
20. A sequence a 0 , a 1 , a 2 , . . . is defined recursively as follows:
∗
a 0 = 0;
1
2
for every n ∈ N, a n+1 = (a n ) + .
4
Prove that for all n ≥ 1, 0 < a n < 1.
21. Explain the paradox in the proof of Theorem 6.3.4, in which we made the
proof easier by changing the goal to a statement that looked like it would
be harder to prove.
6.4. Strong Induction

In the induction step of a proof by mathematical induction, we prove that a
natural number has some property based on the assumption that the previous
number has the same property. In some cases this assumption isn’t strong