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Proof by Mathematical Induction 261
We’ll say more about the justification of the method of mathematical induc-
tion later, but first let’s look at an example of a proof that uses mathematical
induction. The following list of calculations suggests a surprising pattern:
0
1
2 = 1 = 2 − 1
0
2
1
2 + 2 = 1 + 2 = 3 = 2 − 1
2
0
1
3
2 + 2 + 2 = 1 + 2 + 4 = 7 = 2 − 1
4
0
2
3
1
2 + 2 + 2 + 2 = 1 + 2 + 4 + 8 = 15 = 2 − 1
The general pattern appears to be:
1
n
0
2 + 2 +· · · + 2 = 2 n+1 − 1.
Will this pattern hold for all values of n? Let’s see if we can prove it.
0 1 n
Example 6.1.1. Prove that for every natural number n, 2 + 2 +· · · + 2 =
2 n+1 − 1.
Scratch work
Our goal is to prove the statement ∀n ∈ NP(n), where P(n) is the statement
n
1
0
2 + 2 + ··· + 2 = 2 n+1 − 1. According to our strategy, we can do this by
proving two other statements, P(0) and ∀n ∈ N(P(n) → P(n + 1)).
0
1
Plugging in 0 for n, we see that P(0) is simply the statement 2 = 2 − 1,
the first statement in our list of calculations. The proof of this is easy – just do
the arithmetic to verify that both sides are equal to 1. Often the base case of an
induction proof is very easy, and the only hard work in figuring out the proof
is in carrying out the induction step.
For the induction step, we must prove ∀n ∈ N(P(n) → P(n + 1)). Of
course, all of the proof techniques discussed in Chapter 3 can be used in
mathematical induction proofs, so we can do this by letting n be an arbitrary
natural number, assuming that P(n) is true, and then proving that P(n + 1) is
true. In other words, we’ll let n be an arbitrary natural number, assume that
1
1
0
0
n
2 + 2 + ··· + 2 = 2 n+1 − 1, and then prove that 2 + 2 +· · · + 2 n+1 =
2 n+2 − 1. This gives us the following givens and goal:
Givens Goal
1
0
n ∈ N 2 + 2 +· · · + 2 n+1 = 2 n+2 − 1
0
1
n
2 + 2 +· · · + 2 = 2 n+1 − 1
Clearly the second given is similar to the goal. Is there some way to start
with the second given and derive the goal using algebraic steps? The key to the

