Page 279 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proof by Mathematical Induction 265
The base case n = 5 has already been checked in the table. For the induc-
2
n
tion step, we let n ≥ 5 be arbitrary, assume 2 > n , and try to prove that
2
2 n+1 > (n + 1) . How can we relate the inductive hypothesis to the goal?
Perhaps the simplest relationship involves the left sides of the two inequal-
n
ities: 2 n+1 = 2 · 2 . Thus, multiplying both sides of the inductive hypothesis
2
n
2
2 > n by 2, we can conclude that 2 n+1 > 2n . Now compare this inequality
2
2
2
to the goal, 2 n+1 > (n + 1) . If we could prove that 2n ≥ (n + 1) , then the
goal would follow easily. So let’s forget about the original goal and see if we
2
2
can prove that 2n ≥ (n + 1) .
Multiplying out the right side of the new goal we see that we must prove that
2
2
2
2n ≥ n + 2n + 1, or in other words n ≥ 2n + 1. This isn’t hard to prove:
2
Since we’ve assumed that n ≥ 5, it follows that n ≥ 5n = 2n + 3n > 2n + 1.
Solution
n
2
Theorem. For every natural number n ≥ 5, 2 > n .
Proof. By mathematical induction.
n
2
Base case: When n = 5wehave2 = 32 > 25 = n .
n
2
Induction step: Let n ≥ 5 be arbitrary, and suppose that 2 > n . Then
2 n+1 = 2 · 2 n
> 2n 2 (by inductive hypothesis)
2
= n + n 2
2
≥ n + 5n (since n ≥ 5)
2
= n + 2n + 3n

2
2
> n + 2n + 1 = (n + 1) .
Exercises
1. Prove that for all n ∈ N, 0 + 1 + 2 +· · · + n = n(n + 1)/2.
∗
2
2
2
2
2. Prove that for all n ∈ N, 0 + 1 + 2 +· · · + n = n(n + 1)(2n + 1)/6.
3. Prove that for all n ∈ N, 0 + 1 + 2 +· · · + n = [n(n + 1)/2] .
∗ 3 3 3 3 2
4. Find a formula for 1 + 3 + 5 +· · · + (2n − 1), for n ≥ 1, and prove that
your formula is correct. (Hint: First try some particular values of n and
look for a pattern.)
5. Prove that for all n ∈ N,0 · 1 + 1 · 2 + 2 · 3 +· · · + n(n + 1) =
n(n + 1)(n + 2)/3.
6. Find a formula for 0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 +· · · + n(n + 1)(n + 2),
for n ∈ N, and prove that your formula is correct. (Hint: Compare this
exercise to exercises 1 and 5, and try to guess the formula.)