Page 298 - HOW TO PROVE IT: A Structured Approach, Second Edition

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284 Mathematical Induction
Scratch work
It’s probably a good idea to start out by computing the first few terms in the
sequence. We already know a 0 = 0, so plugging in n = 0 in the second equation
we get a 1 = 2a 0 + 1 = 0 + 1 = 1. Thus, plugging in n = 1, we get a 2 = 2a 1 +
1 = 2 + 1 = 3. Continuing in this way we get the following table of values.
n a n
0 0
1 1
2 3
3 7
4 15
5 31
6 63
. .
. . . .
Aha! The numbers we’re getting are one less than the powers of 2. It looks
n
like the formula is probably a n = 2 − 1, but we can’t be sure this is right unless
we prove it. Fortunately, it is fairly easy to prove the formula by induction.

Solution
Theorem. If the sequence a 0 , a 1 , a 2 ,... is defined by the recursive definition
n
given earlier, then for every natural number n, a n = 2 − 1.
Proof. By induction.
0
Base case: a 0 = 0 = 2 − 1.
n
Induction step: Suppose a n = 2 − 1. Then
a n+1 = 2a n + 1 (definition of a n+1 )
n
= 2(2 − 1) + 1 (inductive hypothesis)
= 2 n+1 − 2 + 1 = 2 n+1 − 1.
We end this section with a rather unusual example. We’ll prove that for every
n
real number x > −1 and every natural number n, (1 + x) > nx. A natural way
to proceed would be to let x > −1 be arbitrary, and then use induction on n.
n
In the induction step we assume that (1 + x) > nx, and then try to prove that
(1 + x) n+1 > (n + 1)x. Because we’ve assumed x > −1, we have 1 + x > 0,
n
so we can multiply both sides of the inductive hypothesis (1 + x) > nx by
1 + x to get
(1 + x) n+1 = (1 + x)(1 + x) n
> (1 + x)nx
2
= nx + nx .