Page 299 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Recursion 285
But the conclusion we need for the induction step is (1 + x) n+1 > (n + 1)x,
and it’s not clear how to get this conclusion from the inequality we’ve
derived.
Our solution to this difficulty will be to replace our original problem with a
problem that appears to be harder but is actually easier. Instead of proving the
n
n
inequality (1 + x) > nx directly, we’ll prove (1 + x) ≥ 1 + nx, and then
n
observe that since 1 + nx > nx, it follows immediately that (1 + x) > nx.
n
You might think that if we had difficulty proving (1 + x) > nx, we’ll surely
n
have more difficulty proving the stronger statement (1 + x) ≥ 1 + nx. But it
turns out that the approach we tried unsuccessfully on the original problem
works perfectly on the new problem!
n
Theorem 6.3.4. Forevery x > −1andeverynaturalnumbern,(1 + x) > nx.
Proof. Let x > −1 be arbitrary. We will prove by induction that for ev-
n
ery natural number n, (1 + x) ≥ 1 + nx, from which it clearly follows that
n
(1 + x) > nx.
0
n
Base case: If n = 0, then (1 + x) = (1 + x) = 1 = 1 + 0 = 1 + nx.
n
Induction step: Suppose (1 + x) ≥ 1 + nx. Then
(1 + x) n+1 = (1 + x)(1 + x) n
≥ (1 + x)(1 + nx) (by inductive hypothesis)
= 1 + x + nx + nx 2
2
≥ 1 + (n + 1)x (since nx ≥ 0).

Exercises

n 1

1. Find a formula for and prove that your formula is correct.
∗
i=1 i(i+1)
2. Prove that for all n ≥ 1,
n 2
1 n + 3n
= .
i(i + 1)(i + 2) 4(n + 1)(n + 2)
i=1
3. Prove that for all n ≥ 2,
n 2
1 3n − n − 2
= .
(i − 1)(i + 1) 4n(n + 1)
i=2
4. Prove that for all n ∈ N,
n
(n + 1)(2n + 1)(2n + 3)
2
(2i + 1) = .
3
i=0