1. Section 1.3 - Problem 4
We see that
1
1·2
=
1
2
and
1
1·2
+
1
2·3
=
n
X
k=1
2
3
and
1
1·2
+
1
2·3
+
1
3·4
=
3
4
and conjecture that
1
n
=
k(k + 1)
n+1
This is true for the base case, where n = 1, as shown above.
For the inductive step we assume the conjecture is true for n and prove for n + 1.
n+1
X
k=1
n
X
1
1
1
=
+
k(k + 1) k=1 k(k + 1) (n + 1)(n + 2)
1
n
+
, by the induction assumption
n + 1 (n + 1)(n + 2)
n(n + 2) + 1
=
(n + 1)(n + 2)
n2 + 2n + 1
=
(n + 1)(n + 2)
(n + 1)2
=
(n + 1)(n + 2)
n+1
.
=
n+2
P
1
n
Thus by induction we have nk=1 k(k+1)
= n+1
for all n ∈ Z, n ≥ 1.
=
2. Section 1.3 - Problem 8
P
The base case is 1k=1 j 3 = 1 = [1(1 + 1)/2]2 , so the result holds for n = 1.
For the inductive step we prove the result is true for n + 1 if we assume the result is true for
n. We have
n
n+1
X
X
j 3 + (n + 1)3
j3 =
k=1
k=1
= [n(n + 1)/2]2 + (n + 1)3 , by the induction assumption
n2
+ n + 1)
4
n2 + 4n + 4
= (n + 1)2
4
2
2 (n + 2)
= (n + 1)
4
2
(n + 1)(n + 2)
=
.
2
P
Thus by induction we have nk=1 j 3 = [n(n + 1)/2]2 for all n ∈ Z, n ≥ 1.
= (n + 1)2 (
3. Section 1.3 - Problem 12
P
The base case is 1k=1 j · j! = 1 · 1! = 1 = 2! − 1, so the result holds for n = 1.
For the inductive step we prove the result is true for n + 1 if we assume the result is true for
n. We have
n+1
X
k=1
j · j! =
n
X
j · j! + (n + 1) · (n + 1)!
k=1
= (n + 1)! − 1 + (n + 1) · (n + 1)! , by the induction assumption
= (n + 1 + 1) · (n + 1)! − 1
= (n + 2)! − 1
P
Thus by induction we have nk=1 j · j! = (n + 1)! − 1 for all n ∈ Z, n ≥ 1.
4. Section 1.3 - Problem 14
The problem here is equivalent to showing that for all integers n ≥ 54, there exist nonnegative integers a and b such that n = 7a + 10b.
Method 1
The base case is shown to be true, as illustrated by 54 = 7 · 2 + 10 · 4.
For the inductive step we assume that the result is true for some n ≥ 54, and we prove the
result for n + 1. By the inductive assumption we can write n = 7a + 10b.
Suppose b ≥ 2. Then as 21 = 3 · 7 = 2 · 10 + 1, we have
n + 1 = 7(a + 3) + 10(b − 2)
Otherwise we have b ≤ 1. As n = 7a + 10b ≥ 54, we thus have 7a ≥ n − 10 ≥ 44 and so
a ≥ 7. As 50 = 5 · 10 = 7 · 7 + 1 we then have
n + 1 = 7(a − 7) + 10(b + 5)
In both cases we can write n + 1 = 7c + 10d for some non-negative integers c and d. Thus by
induction our result holds for all n ≥ 54.
Method 2
An alternative approach uses strong induction. We first show that the result holds for the
following integers:
54 = 7 · 2 + 10 · 4
55 = 7 · 5 + 10 · 2
56 = 7 · 8 + 10 · 0
57 = 7 · 1 + 10 · 5
58 = 7 · 4 + 10 · 3
59 = 7 · 7 + 10 · 1
60 = 7 · 0 + 10 · 6
For the inductive step, let n ≥ 60 and assume the result holds for all k such that 54 ≤ k ≤ n.
We prove the result holds for n + 1.
As n ≥ 60 we have n − 6 ≥ 54, and by the inductive assumption we can write n − 6 =
7a + 10b. Then n + 1 = 7(a + 1) + 10b. Thus by strong induction our result holds for all
n ≥ 54.
5. Section 1.3 - Problem 22
For the base case the inequality becomes 1 + 0h = 1 = (1 + h)0 , which is true, so the result
holds for n = 0.
For the inductive step we prove the result is true for n + 1 if we assume the result is true for
n. We have
(1 + h)n+1 = (1 + h)n (1 + h)
≥ (1 + nh)(1 + h) , by the induction assumption, and as 1 + h ≥ 0
= 1 + h + nh + nh2
≥ 1 + h + nh
= 1 + (n + 1)h.
Thus by induction we have 1 + nh ≤ (1 + h)n for all n ∈ Z, n ≥ 0.
6. Section 1.3 - Problem 24
The proof fails in the induction argument. It reasons that the sets {1, . . . , n} and
{2, . . . , n + 1} will have common members. This is false when n + 1 = 2. In this case the sets
are {1} and {2}, which are disjoint, and the argument breaks down.
7. Section 1.3 - Problem 34
The base step here is clear. In a 2 × 2 chessboard with one square missing, the 3 other squares
form an L-shaped piece.
Assume that any 2n × 2n chessboard with one square missing can be filled by L-shaped
pieces, and consider a 2n+1 × 2n+1 chessboard with an arbitrary square missing. This 2n+1 ×
2n+1 chessboard consists of 4 2n × 2n chessboard arranged in a square. Exactly one of these
smaller chessboards will be missing one square, so by the inductive assumption, this smaller
chessboard can be covered using L-shaped pieces. The other 3 smaller chessboards have no
squares missing. Place one L-shaped piece in the center of the larger piece, so that 1 square of
this L-shaped piece lies in each of these 3 smaller chessboards. Now each of these 3 smaller
chessboards need to have all of their pieces except this 1 at the center covered by L-shaped
pieces. By the inductive assumption, this is possible.
Hence by induction, any 2n × 2n chessboard with one square missing can be covered by
L-shaped pieces.
8. Section 3.1 - Problem 2
The numbers
(a) 201 = 3 · 67, (b) 203 = 7 · 29, (c) 207 = 3 · 69, (e) 213 = 3 · 71, and (f) 221 = 13 · 17
are all composite. The only prime√is (d) 211. This can be seen by checking that 211 is not
divisible by any of the primes p ≤ 203 < 15, i.e., by p = 2, 3, 5, 7, 11 and 13.
9. Section 3.1 - Problem 6
We factor n3 + 1 = (n + 1)(n2 − n + 1). Suppose n3 + 1 is prime. Then n3 + 1 > 1 and so
n > 0. Hence n + 1 > 1 and so by the factoring above we require n + 1 = n3 + 1. This equality
holds if and only if n = 1, in which case the number is 13 + 1 = 2.
In conclusion, no integer of the form n3 + 1 is prime, other then 2 = 13 + 1.
10. Section 3.1 - Problem 8
As Qn = n! + 1 > 1, there exists a prime p dividing Qn . Suppose p ≤ n. Then p divides n!
and therefore p divides Qn − n! = 1, which is impossible. Thus p > n.
Suppose there are finitely many primes. Then there exists a largest prime n. But by the above
argument, there exists a prime p > n dividing n! + 1. This is a contradiction and thus there are
infinitely many primes.