Page 280 - HOW TO PROVE IT: A Structured Approach, Second Edition

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266 Mathematical Induction
7. Find a formula for 3 + 3 + 3 +· · · + 3 , for n ≥ 0, and prove that your
∗ 0 1 2 n
formula is correct. (Hint: Try to guess the formula, basing your guess on
Example 6.1.1. Then try out some values of n and adjust your guess if
necessary.)
8. Prove that for all n ≥ 1,

1 1 1 1 1
1 − + − +· · · + −
2 3 4 2n − 1 2n
1 1 1 1
= + + + ··· +
n + 1 n + 2 n + 3 2n
2
9. (a) Prove that all n ∈ N, 2 | (n + n).
3
(b) Prove that for all n ∈ N, 6 | (n − n).
∗ n
10. Prove that for all n ∈ N, 64 | (9 − 8n − 1).
n
11. Prove that for all n ∈ N,9 | (4 + 6n − 1).
n
n
12. Prove that for all integers a and b and all n ∈ N, (a − b) | (a − b ).
(Hint: Let a and b be arbitrary integers and then prove by induction
n
n
that ∀n ∈ N[(a − b) | (a − b )]. For the induction step, you must relate
n
n
a n+1 − b n+1 to a − b . You might find it useful to start by completing
n
n
the following equation: a n+1 − b n+1 = a(a − b ) + ? .)
13. Prove that for all integers a and b and all n ∈ N,(a + b) | (a 2n+1 + b 2n+1 ).
3
n
∗ 14. Prove that for all n ≥ 10, 2 > n .
15. Prove that for all n ∈ N, either n ≡ 0 (mod 3) or n ≡ 1 (mod 3) or n ≡ 2
(mod 3). (Recall that this notation was introduced in Definition 4.6.9.)
1 2 3 n
16. Prove that for all n ≥ 1, 2 · 2 + 3 · 2 + 4 · 2 +· · · + (n + 1)2 =
n2 n+1 .
0
17. (a) What’s wrong with the following proof that for all n ∈ N,1 · 3 +
1
2
n
3 · 3 + 5 · 3 +· · · + (2n + 1)3 = n3 n+1 ?
Proof. We use mathematical induction. Let n be an arbitrary
0 1 2
natural number, and suppose that 1 · 3 + 3 · 3 + 5 · 3 +· · · +
n
(2n + 1)3 = n3 n+1 . Then
n
0
1
2
1 · 3 + 3 · 3 + 5 · 3 +· · · + (2n + 1)3 + (2n + 3)3 n+1
= n3 n+1 + (2n + 3)3 n+1
= (3n + 3)3 n+1
= (n + 1)3 n+2 ,
as required.
n
1
2
0
(b) Find a formula for 1 · 3 + 3 · 3 + 5 · 3 + ··· + (2n + 1)3 , and
prove that your formula is correct.