Page 368 - HOW TO PROVE IT: A Structured Approach, Second Edition

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354 Appendix 1: Solutions to Selected Exercises
3. Base case: When n = 0, both sides of the equation are 0.
3 3 3 3
Induction step: Suppose that n ∈ N and 0 + 1 + 2 +· · · + n =
2
[n(n + 1)/2] . Then
3
3
3
3
3
3
3
3
0 + 1 + 2 +· · · + (n + 1) = (0 + 1 + 2 +· · · + n ) + (n + 1) 3
n(n + 1) 3

2
= + (n + 1)
2
n

2
2
= (n + 1) + n + 1
4
2
n + 4n + 4
2
= (n + 1) ·
4

2
(n + 1)(n + 2)
= .
2
7. Hint: The formula is (3 n+1 − 1)/2.
n
n
10. Base case: When n = 0, 9 − 8n − 1 = 0 = 64 · 0, so 64 | (9 − 8n − 1).
n
Induction step: Suppose that n ∈ N and 64 | (9 − 8n − 1). Then there
n
is some integer k such that 9 − 8n − 1 = 64k. Therefore
9 n+1 − 8(n + 1) − 1 = 9 n+1 − 8n − 9
= 9 n+1 − 72n − 9 + 64n
n
= 9(9 − 8n − 1) + 64n
= 9(64k) + 64n
= 64(9k + n),
so 64 | (9 n+1 − 8(n + 1) − 1).
n
3
14. Base case: When n = 10, 2 = 1024 > 1000 = n .
n
3
Induction step: Suppose n ≥ 10 and 2 > n . Then
2 n+1 = 2 · 2 n
> 2n 3 (by inductive hypothesis)
3
= n + n 3
3
≥ n + 10n 2 (since n ≥ 10)
2
3
= n + 3n + 7n 2
3
2
≥ n + 3n + 70n (since n ≥ 10)
2
3
= n + 3n + 3n + 67n
3
2
3
> n + 3n + 3n + 1 = (n + 1) .
19. (a) Basecase:Whenn = 1,thestatementtobeprovenis0 < a < b,which
was given.