46
Chapter 4 Mathematical Induction
Section 4.1. The Principle of Mathematical Induction
1. Proof. We proceed by induction. Since
1
2·3
=
1
2·1+4 ,
the formula holds for n = 1. Assume that
1
1
1
k
+
+ ···+
=
2·3 3·4
(k + 1)(k + 2)
2k + 4
where k is a positive integer. We show that
1
1
1
k+1
+
+ ···+
=
.
2·3 3·4
(k + 2)(k + 3)
2k + 6
Observe that
1
1
1
+
+ ··· +
2·3
3·4
(k + 2)(k + 3)
1
1
1
+
+ ··· +
2·3
3·4
(k + 1)(k + 2)
–
1
(k + 2)(k + 3)
=
»
=
k
1
k(k + 3) + 2
+
=
2k + 4
(k + 2)(k + 3)
2(k + 2)(k + 3)
=
(k + 1)(k + 2)
k+1
k2 + 3k + 2
=
=
.
2(k + 2)(k + 3)
2(k + 2)(k + 3)
2k + 6
+
By the Principle of Mathematical Induction,
1
1
1
n
+
+ ···+
=
2·3 3·4
(n + 1)(n + 2)
2n + 4
for every positive integer n.
2. Proof. We verify this formula by mathematical induction. Since
holds for n = 1. Assume that
1
1·3
=
1
2·1+1
= 13 , the formula
1
1
1
1
k
+
+
+ ···+
=
1·3 3·5 5·7
(2k − 1)(2k + 1)
2k + 1
for some positive integer k. We show that
1
1
1
k+1
1
+
+
+ ···+
=
.
1·3 3·5 5·7
(2(k + 1) − 1)(2(k + 1) + 1)
2k + 3
Observe that
and
1
1
=
(2(k + 1) − 1)(2(k + 1) + 1)
(2k + 1)(2k + 3)
1
1
1
1
+
+
+ ···+
1·3
3·5
5·7
(2k + 1)(2k + 3)
1
1
1
1
+
+
+ ··· +
1·3
3·5
5·7
(2k − 1)(2k + 1)
1
+
(2k + 1)(2k + 3)
=
»
=
1
k(2k + 3) + 1
k
+
=
2k + 1
(2k + 1)(2k + 3)
(2k + 1)(2k + 3)
=
2k2 + 3k + 1
(k + 1)(2k + 1)
k+1
=
=
.
(2k + 1)(2k + 3)
(2k + 1)(2k + 3)
2k + 3
By the Principle of Mathematical Induction,
1
1
1
n
1
+
+
+ ···+
=
1·3 3·5 5·7
(2n − 1)(2n + 1)
2n + 1
for every positive integer n.
–
47
3. Proof. We proceed by induction. Since
1
1·4
=
1
3+1 ,
the formula holds for n = 1. Assume that
1
1
1
k
+
+ ···+
=
1·4 4·7
(3k − 2)(3k + 1)
3k + 1
where k is an arbitrary positive integer. We show that
1
1
k+1
1
+
+ ···+
=
.
1·4 4·7
(3k + 1)(3k + 4)
3k + 4
Observe that
1
1
1
+
+ ··· +
1·4
4·7
(3k + 1)(3k + 4)
1
1
1
+
+ ··· +
1·4
4·7
(3k − 2)(3k + 1)
–
1
(3k + 1)(3k + 4)
=
»
=
k
1
k(3k + 4) + 1
+
=
3k + 1
(3k + 1)(3k + 4)
(3k + 1)(3k + 4)
=
(k + 1)(3k + 1)
k+1
3k2 + 4k + 1
=
=
.
(3k + 1)(3k + 4)
(3k + 1)(3k + 4)
3k + 4
+
By the Principle of Mathematical Induction,
1
1
1
n
+
+ ···+
=
1·4 4·7
(3n − 2)(3n + 1)
3n + 1
for every positive integer n.
4. Claim: For every positive integer n,
2 + 4 + 6 + · · · + 2n = n(n + 1).
Proof. We proceed by induction. Since 2 = 1(1 + 1), the statement is true for n = 1. Assume
that 2+4+6+· · ·+2k = k(k+1) for a positive integer k. We show that 2+4+6+· · ·+2(k+1) =
(k + 1)(k + 2). Observe that
2 + 4 + 6 + · · · + 2(k + 1) = [2 + 4 + 6 + · · · + 2k] + 2(k + 1)
= k(k + 1) + 2(k + 1) = (k + 1)(k + 2).
By the Principle of Mathematical Induction, 2 + 4 + 6 + · · · + 2n = n(n + 1) for every positive
integer n.
5. Proof. We proceed by induction. Since 1 = 1(2 − 1), the formula holds for n = 1. Assume
that
1 + 5 + 9 + · · · + (4k − 3) = k(2k − 1)
for an arbitrary positive integer k. We show that
1 + 5 + 9 + · · · + (4k + 1) = (k + 1)(2k + 1).
Now
1 + 5 + 9 + · · · + (4k + 1) =
=
=
[1 + 5 + 9 + · · · + (4k − 3)] + (4k + 1)
k(2k − 1) + (4k + 1) = 2k 2 + 3k + 1
(k + 1)(2k + 1).
By the Principle of Mathematical Induction,
1 + 5 + 9 + · · · + (4n − 3) = n(2n − 1)
for every positive integer n.
48
6. Proof. We proceed by induction. Since 2 = 1 · (3 · 1 + 1)/2, the statement is true for n = 1.
Assume that 2 + 5 + 8 + · · · + (3k − 1) = k(3k + 1)/2 for a positive integer k. We show that
2 + 5 + 8 + · · · + [3(k + 1) − 1] = (k + 1)[3(k + 1) + 1]/2 = (k + 1)(3k + 4)/2. Observe that
2 + 5 + 8 + · · · + [3(k + 1) − 1] =
[2 + 5 + 8 + · · · + (3k − 1)] + (3k + 2)
k(3k + 1)
k(3k + 1) + 2(3k + 2)
=
+ (3k + 2) =
2
2
(k + 1)(3k + 4)
3k 2 + 7k + 4
=
=
.
2
2
By the Principle of Mathematical Induction, 2 + 5 + 8 + · · · + (3n − 1) = n(3n + 1)/2 for every
positive integer n.
7. Proof. We proceed by induction. Since 12 =
1·2·3
6 ,
the formula holds for n = 1. Assume that
k(k + 1)(2k + 1)
6
for an arbitrary positive integer k. We show that
12 + 22 + · · · + k 2 =
12 + 22 + · · · + (k + 1)2 =
(k + 1)(k + 2)(2k + 3)
.
6
We have
12 + 22 + · · · + (k + 1)2
! 2
"
1 + 22 + · · · + k 2 + (k + 1)2
=
k(k + 1)(2k + 1)
+ (k + 1)2
6
k(k + 1)(2k + 1) + 6(k + 1)2
6
(k + 1)(2k 2 + 7k + 6)
6
(k + 1)(k + 2)(2k + 3)
.
6
=
=
=
=
By the Principle of Mathematical Induction,
1 2 + 2 2 + · · · + n2 =
n(n + 1)(2n + 1)
6
for every positive integer n.
8. Proof. We verify this formula by mathematical induction. Since 13 =
3
3
3
3
k2 (k+1)2
4
12 (1+1)2
4
= 1, the
for a positive integer
formula holds for n = 1. Assume that 1 + 2 + 3 + · · · + k =
k. We show that
(k + 1)2 (k + 2)2
13 + 23 + 33 + · · · + (k + 1)3 =
.
4
Observe that
!
"
13 + 23 + 33 + · · · + (k + 1)3 = 13 + 23 + 33 + · · · + k 3 + (k + 1)3
k 2 (k + 1)2
k 2 (k + 1)2 + 4(k + 1)3
+ (k + 1)3 =
4
4
(k + 1)2 (k 2 + 4k + 4)
(k + 1)2 (k + 2)2
=
=
.
4
4
By the Principle of Mathematical Induction,
=
1 3 + 2 3 + 3 3 + · · · + n3 =
for every positive integer n.
n2 (n + 1)2
4
49
9. Proof. We use induction. Since 1 + 3 = 4 =
32 −1
2 ,
the formula holds for n = 1. Assume that
1 + 3 + 32 + · · · + 3k =
3k+1 − 1
2
for an arbitrary positive integer k. We show that
1 + 3 + 32 + · · · + 3k+1 =
3k+2 − 1
.
2
Observe that
1 + 3 + 32 + · · · + 3k+1
!
"
1 + 3 + 32 + · · · + 3k + 3k+1
=
3k+1 − 1 + 2 · 3k+1
3k+1 − 1
+ 3k+1 =
2
2
3k+2 − 1
3 · 3k+1 − 1
=
.
2
2
=
=
By the Principle of Mathematical Induction,
1 + 3 + 32 + · · · + 3n =
3n+1 − 1
2
for every positive integer n.
10. Proof. We use induction. Since
r2 − 1
(r + 1)(r − 1)
=
= 1 + r,
r−1
r−1
the formula holds for n = 1. Assume that
1 + r + r2 + · · · + rk =
rk+1 − 1
r−1
for a positive integer k. We show that
1 + r + r2 + · · · + rk+1 =
rk+2 − 1
.
r−1
Observe that
1 + r + r2 + · · · + rk+1
=
=
=
(1 + r + r2 + · · · + rk ) + rk+1
(rk+1 − 1) + rk+1 (r − 1)
rk+1 − 1
+ rk+1 =
r−1
r−1
rk+1 − 1 + rk+2 − rk+1
rk+2 − 1
=
.
r−1
r−1
By the Principle of Mathematical Induction,
1 + r + r2 + · · · + rn =
rn+1 − 1
r−1
for every positive integer n.
11. Proof. We use induction. Since 1 · 2 =
1(1+1)(1+2)
,
3
the formula holds for n = 1. Assume that
1 · 2 + 2 · 3 + 3 · 4 + · · · + k(k + 1) =
k(k + 1)(k + 2)
3
50
for a positive integer k. We show that
1 · 2 + 2 · 3 + 3 · 4 + · · · + (k + 1)(k + 2) =
(k + 1)(k + 2)(k + 3)
.
3
Observe that
1 · 2 + 2 · 3 + 3 · 4 + · · · + (k + 1)(k + 2)
= [1 · 2 + 2 · 3 + 3 · 4 + · · · + k(k + 1)] + (k + 1)(k + 2)
k(k + 1)(k + 2)
+ (k + 1)(k + 2)
=
3
k(k + 1)(k + 2) + 3(k + 1)(k + 2)
=
3
(k + 1)(k + 2)(k + 3)
= =
.
3
By the Principle of Mathematical Induction,
1 · 2 + 2 · 3 + 3 · 4 + · · · + n(n + 1) =
n(n + 1)(n + 2)
3
for every positive integer n.
12. (a) Proof. Assume that k 2 + k + 5 is even. Then k 2 + k + 5 = 2x for some integer x. Then
(k + 1)2 + (k + 1) + 5
= k 2 + 2k + 1 + k + 1 + 5
= (k 2 + k + 5) + (2k + 2)
= 2x + 2k + 2 = 2(x + k + 1).
Since x + k + 1 is an integer, (k + 1)2 + (k + 1) + 5 is even.
(b) When k = 7, k 2 + k + 5 = 61.
(c) The integer n2 + n + 5 is not even for every positive integer n. Part (a) is the inductive
step in an induction proof to prove that n2 + n + 5 is even for every integer n. Part (b)
shows that n2 + n + 5 is not even for every positive integer n.
13. (a) For n = 1, (n + 2)(n + 3)/2 = 6, which is even.
(b) Proof. Assume that (k + 2)(k + 3)/2 is even where k ∈ N. Then (k + 2)(k + 3)/2 = 2x
2
= 2x. Hence
for some integer x. Thus k +5k+6
2
(k + 3)(k + 4)
2
=
=
k 2 + 7k + 12
(k 2 + 5k + 6) + (2k + 6)
=
2
2
k 2 + 5k + 6
+ k + 3 = 2x + k + 3.
2
This is not even when k = 2, say.
(c) When k = 2, (k + 2)(k + 3)/2 = 10 and (k + 3)(k + 4)/2 = 15. Hence the statement in (b)
is false, which is the induction step in an induction proof which attempts to prove that
(n + 2)(n + 3)/2 is even for every positive integer n.
14. (a) 11 − 1 = 10.
(b) 11 − 2 = 9.
(c) 11 − 10 = 1.
51
(d) If we remove one tile from each square, we have removed 100 = 102 tiles. Now each
square in s10 has no tile and each square in sk (1 ≤ k ≤ 9) contains 10 − k tiles. If we
remove one tile from each square in the strips s1 , s2 , . . . , s9 , we have removed 92 tiles.
Continuing in this manner, we see that the total number of tiles by Result 4.5 is
12 + 22 + · · · + 102 =
10 · 11 · 21
= 385.
6
(e) For each positive integer n,
n
!
k2 =
k=1
n
!
k=1
(n + 1 − k)(2k − 1) =
n(n + 1)(2n + 1)
.
6
Section 4.2. Additional Examples of Induction Proofs
1. Proof. We proceed by induction. The inequality holds for n = 0 since 20 = 1 > 0. Assume
that 2k > k, where k is a nonnegative integer. We show that 2k+1 > k + 1. When k = 0, we
have 2k+1 = 2 > 1 = k + 1. We therefore assume that k ≥ 1. Then
2k+1 = 2 · 2k > 2k = k + k ≥ k + 1.
By the Principle of Mathematical Induction, 2n > n for every nonnegative integer n.
2. Proof. We proceed by induction. Since 7! = 5040 > 2187 = 37 , the inequality holds for n = 7.
Suppose that k! > 3k for an integer k ≥ 7. We show that (k + 1)! > 3k+1 . Now
(k + 1)! = (k + 1)k! > (k + 1)3k ≥ 8 · 3k > 3 · 3k = 3k+1 .
By the Principle of Mathematical Induction, n! > 3n for every integer n ≥ 7.
3. Proof. We proceed by induction. Since 4! = 24 > 16 = 42 , the inequality holds for n = 4.
Assume that k! > k 2 for an arbitrary integer k ≥ 4. We show that (k + 1)! > (k + 1)2 . Observe
that
(k + 1)! =
=
>
(k + 1)k! > (k + 1)k 2 = k 3 + k 2
k 2 + k · k 2 ≥ k 2 + 16k = k 2 + 2k + 14k
k 2 + 2k + 1 = (k + 1)2 .
By the Principle of Mathematical Induction, n! > n2 for every integer n ≥ 4.
4. Proof. We proceed by induction. Since 31 > 1, the statement is true for n = 1. Assume that
3k > k for an arbitrary positive integer k. We show that 3k+1 > k + 1. Observe that
3k+1 = 3 · 3k > 3k = k + 2k ≥ k + 2 > k + 1.
By the Principle of Mathematical Induction, 3n > n for every positive integer n.
5. Proof. We proceed by induction. Since n2 = 2n + 3 for n = 3, the inequality holds for n = 3.
Assume that k 2 ≥ 2k + 3 for an integer k ≥ 3. We show that (k + 1)2 ≥ 2k + 5. Now
(k + 1)2
=
≥
k 2 + 2k + 1 ≥ (2k + 3) + (2k + 1)
(2k + 3) + 7 = 2k + 10 > 2k + 5.
By the Principle of Mathematical Induction, n2 ≥ 2n + 3 for every integer n ≥ 3.