55
for some positive integer k. We show that
2! · 4! · 6! · · · (2k + 2)! ≥ ((k + 2)!)k+1 .
Observe that
2! · 4! · 6! · (2k + 2)! =
≥
=
≥
=
[2! · 4! · 6! · · · (2k)!] (2k + 2)!
((k + 1)!)k (2k + 2)!
((k + 1)!)k (k + 1)! [(2k + 2)(2k + 1) · · · (k + 2)]
((k + 1)!)k+1 [(k + 2)(k + 2) · · · (k + 2)]
((k + 1)!)k+1 (k + 2)k+1 = ((k + 2)!)k+1 .
By the Principle of Mathematical Induction, the inequality holds for every positive integer n.
17. Proof. We proceed by induction. Since the sum of the interior angles of each triangle is
180o = (3 − 2) · 180o , the result holds for n = 3. Assume that the sum of the interior angles
of every k-gon is (k − 2) · 180o for an arbitrary integer k ≥ 3. We show that the sum of the
interior angles of every (k + 1)-gon is (k − 1) · 180o. Let Pk+1 be a (k + 1)-gon whose k + 1
vertices are v1 , v2 , . . ., vk+1 and whose edges are v1 v2 , v2 v3 , . . ., vk vk+1 , vk+1 v1 . Now let Pk
be the k-gon whose vertices are v1 , v2 , . . . , vk and whose edges are v1 v2 , v2 v3 , . . ., vk−1 vk , vk v1
and let P3 be the 3-gon whose vertices are vk , vk+1 , v1 and whose edges are vk vk+1 , vk+1 v1 ,
v1 vk . Observe that the sum of the interior angles of Pk+1 is the sum of interior angles of Pk
and the interior angles of P3 . By the induction hypothesis, the sum of the interior angles of
Pk is (k − 2) · 180o and the sum of the interior angles of P3 is 180o . Therefore, the sum of the
interior angles of Pk+1 is (k − 2) · 180o + 180o = (k − 1) · 180o. By the Principle of Mathematical
Induction, the sum of the interior angles of every n-gon is (n − 2) · 180o.
Section 4.3. Sequences
1. a1 = 5/3, a2 = −7/6, a3 = 3/4, a4 = −11/24.
2. a0 = 0, a1 = 1, a2 = 4/3, a3 = 9/5
3. a1 = −1/22 = −1/4, a2 = 1/24 = 1/16, a3 = −1/26 = −1/64, a4 = 1/256.
4. (a) 1, 1, 2, 2;
(b) 0, 1, 1, 2;
(c) 1, 0, 1, 0.
5. (a) 3, 3, 3, 3
(b) 0, 8, 0, 32
(c) 1, 5, 19, 65.
6. (a) an = (−1)n+1
(d) an = 2n
2
(h) an = 4n − 7
(b) an = 1 + (−1)n
(c) an = 2n − 1
(e) an = 2 · n! (f) an = 2 · 3n
2
(i) an = (n − 1) + 1
2
n −n+2
;
2
7. (1) an = 2n−1 ;
(2) bn =
8. (1) an = 3n−1 ;
(2) bn = 2n2 − 4n + 3;
9. (a) (a, b, c, d, e)
(b) (0, 1, 0, 1, 0, 1)
(g) an = 3n − 2
(j) an = 5
(3) cn = 3n−1 − n! + 1.
(3) cn = 1! + 2! + · · · + n!.
(c) any bit string whose first term is 1 and alternates 0, 1 thereafter.
10. (a) X1 : 10010; X2 : 01011; X3 : 00101
(b) s1 : ∅; s2 : {x2 , x5 }; s3 : X.
11. Each placement of coins corresponds to the 4-bit string a1 a2 a3 a4 , where ai = 1 (1 ≤ i ≤ 4) if
a black coin is placed on square i and ai = 0 if a red coin is placed on square i (or interchange
red and black).
56
12. A1 = {1}, A2 = {2}, A3 = {1, 3}, A4 = {2, 4}, A5 = {1, 3, 5}, A6 = {2, 4, 6}.
13. (a) a2 = 2a1 − 1 = 3; a3 = 5; a4 = 9; a5 = 17.
(b) Result For every positive integer, an = 2n−1 + 1.
Proof. We proceed by induction. Since a1 = 2 = 21−1 + 1, the formula holds for n = 1.
Assume that ak = 2k−1 + 1 for a positive integer k. Now
ak+1 = 2ak − 1 = 2(2k−1 + 1) − 1 = 2k + 1.
By the Principle of Mathematical Induction, an = 2n−1 + 1 for every positive integer n.
14. (a) a2 = 2a1 + 1 = 7, a3 = 15, a4 = 31, a5 = 63.
(b) Result For every positive integer, an = 2n+1 − 1.
Proof. We proceed by induction. Since a1 = 3 = 21+1 − 1, the formula holds for n = 1.
Assume that ak = 2k+1 − 1 for a positive integer k. Now
ak+1 = 2ak + 1 = 2(2k+1 − 1) + 1 = 2k+2 − 1.
By the Principle of Mathematical Induction, an = 2n+1 − 1 for every positive integer n.
15. ai = i for 3 ≤ i ≤ 6.
16. (a) s1 = 2 as both 1-bit strings, 0 and 1, do not contain 1 immediately followed by 0, s2 = 3
as the 2-bit strings 00, 01, 11 do not contain 1 immediately followed by 0, and s3 = 4 as
the 3-bit strings 000, 001, 011, 111 do not contain 1 immediately followed by 0.
(b) s1 = 2 and sn = sn−1 + 1 for n ≥ 2. For n ≥ 2, an n-bit string that does not contain 1
immediately followed by 0 either
(1) has 1 as the last bit and so the first n − 1 bits contain no 1 immediately followed by
0 or
(2) has 0 as the last bit and so all n bits are 0.
Thus sn = sn−1 + 1 for n ≥ 2.
17. (a) s1 = 2 as both 1-bit strings have no three consecutive 0’s, s2 = 4 as all four 2-bit strings
have no three consecutive 0’s, and s3 = 7 as all eight 3-bit strings except 000 have no
three consecutive 0’s.
(b) s1 = 2, s2 = 4, s3 = 7, and sn = sn−1 + sn−2 + sn−3 for n ≥ 4. For n ≥ 4, an n-bit string
with no three consecutive 0’s either
(1) has 1 as the last bit and the first n − 1 bits have no three consecutive 0’s,
(2) has 10 as the last two bits and the first n − 2 bits have no three consecutive 0’s, or
(3) has 100 as the last three bits and the first n − 3 bits have no three consecutive 0’s.
Therefore, sn = sn−1 + sn−2 + sn−3 for n ≥ 4.
18. (a) a4 = 6, a5 = 10 and a6 = 19.
(b) a4 = 6, a5 = 11 and a6 = 20.
(c) a4 = 5, a5 = 6 and a6 = 9.
19. (a) s1 = 0, s2 = 1 and s3 = 3.
(b) s1 = 0. Let S = {a1 , a2 , . . . , an }, where n ≥ 2. The number of 2-element subsets of S
equals the sum of the number of 2-element subsets not containing an and the number of
2-element subsets containing an . Hence sn = sn−1 + (n − 1) for n ≥ 2.
(c) s4 = s3 + (4 − 1) = 3 + 3 = 6.
57
20. (a) a3 = 1, a4 = 1, a5 = 2, a6 = 1, a7 = 1, a8 = 2, a9 = 1.
(b) Proof. Observe that an ∈ {1, 2} for all n ≥ 1. Assume that an−1 an+1 = 1. Then
an−1 = an+1 = 1. Since an+1 = 1, it follows that an an−1 ̸= 1. On the other hand,
an−1 = 1 and so an ̸= 1. Thus an = 2.
21. Proof. We proceed by induction. Since F1 = 1 = 2 − 1 = F3 − 1, the statement is true for
n = 1. Assume that
F1 + F2 + · · · + Fk = Fk+2 − 1
for a positive integer k. We show that
F1 + F2 + · · · + Fk+1 = Fk+3 − 1.
Now
Fk+3 − 1 =
=
Fk+1 + Fk+2 − 1 = (Fk+2 − 1) + Fk+1
F1 + F2 + · · · + Fk + Fk+1 .
By the Principle of Mathematical Induction, F1 + F2 + · · · + Fn = Fn+2 − 1 for every positive
integer n.
22. Proof. We proceed by induction. Since F2 = 1 = 2 − 1 = F3 − 1, the statement is true for
n = 1. Assume that
F2 + F4 + · · · + F2k = F2k+1 − 1
for an arbitrary positive integer k. We show that
F2 + F4 + · · · + F2k+2 = F2k+3 − 1.
Then
F2k+3 − 1
= F2k+1 + F2k+2 − 1 = (F2k+1 − 1) + F2k+2
= F2 + F4 + · · · + F2k + F2k+2 .
By the Principle of Mathematical Induction, F2 + F4 + · · · + F2n = F2n+1 − 1 for every positive
integer n.
23. Proof. We proceed by induction. When n = 2, we have
Fn−1 Fn+2 = F1 F4 = 1 · 3 = 3 and Fn Fn+1 = F2 F3 = 1 · 2 = 2.
Thus when n = 2,
Fn−1 Fn+2 = Fn Fn+1 + (−1)n
is a true statement. Assume that
Fk−1 Fk+2 = Fk Fk+1 + (−1)k
for an integer k ≥ 2. We show that
Fk Fk+3 = Fk+1 Fk+2 + (−1)k+1 .
Observe that
Fk Fk+3
= Fk (Fk+1 + Fk+2 ) = Fk Fk+1 + Fk Fk+2
= Fk−1 Fk+2 − (−1)k + Fk Fk+2
= Fk−1 Fk+2 + Fk Fk+2 + (−1)k+1
= (Fk−1 + Fk )Fk+2 + (−1)k+1 = Fk+1 Fk+2 + (−1)k+1 .
58
By the Principle of Mathematical Induction,
Fn−1 Fn+2 = Fn Fn+1 + (−1)n
for every integer n ≥ 2.
Section 4.4. The Strong Principle of Mathematical Induction
1. (a) a2 = 2, a3 = 4, a4 = 8 and a5 = 16.
(b) Claim: an = 2n−1 for every positive integer n.
(c) Proof. We proceed by induction. Since a1 = 1 = 21−1 = 20 , the formula holds for n = 1.
Assume that ak = 2k−1 for a positive integer k. We show that ak+1 = 2k . Observe that
ak+1 = 2ak = 2 · 2k−1 = 2k .
By the Principle of Mathematical Induction, an = 2n−1 for every positive integer n.
2. (a) a3 = 6, a4 = 12 and a5 = 20.
(b) Claim: an = n(n − 1) for every positive integer n.
(c) Proof. We proceed by induction. Since a1 = 1 · 0 = 0, the formula holds for n = 1.
Assume for an integer k ≥ 1 that ai = i(i − 1) for every integer i with 1 ≤ i ≤ k. We show
that ak+1 = (k + 1)k. When k = 1, we have a2 = 2 = 2 · 1. Therefore, ak+1 = (k + 1)k
when k = 1. When k ≥ 2,
ak+1
= 2ak − ak−1 + 2 = 2k(k − 1) − (k − 1)(k − 2) + 2
= (k − 1)[2k − (k − 2)] + 2 = (k − 1)(k + 2) + 2
= k 2 + k = (k + 1)k.
By the Strong Principle of Mathematical Induction, an = n(n − 1) for every positive
integer n.
3. Proof. We apply the Strong Principle of Mathematical Induction. Since a1 = 4 = 12 + 3, the
formula holds for n = 1. Assume for an arbitrary positive integer k that ai = i2 + 3 for every
integer i with 1 ≤ i ≤ k. We show that ak+1 = (k + 1)2 + 3. Since a2 = 7 = (1 + 1)2 + 3, it
follows that ak+1 = (k + 1)2 + 3 when k = 1. Hence we may assume that k ≥ 2. Observe that
ak+1
= 2ak − ak−1 + 2 = 2(k 2 + 3) − [(k − 1)2 + 3] + 2
= 2k 2 + 6 − (k 2 − 2k + 1 + 3) + 2 = 2k 2 + 6 − k 2 + 2k − 4 + 2
= k 2 + 2k + 4 = (k + 1)2 + 3.
By the Strong Principle of Mathematical Induction, an = n2 + 3 for every positive integer n.
4. Proof. We apply the Strong Principle of Mathematical Induction. Since a1 = 1 = 12 , the
formula holds for n = 1. Assume for an arbitrary positive integer k that ai = i2 for every
integer i with 1 ≤ i ≤ k. We show that ak+1 = (k + 1)2 . Observe that a2 = 4 = 22 and so
ak+1 = (k + 1)2 when k = 1. Hence we may assume that k ≥ 2. Observe that
ak+1
=
=
=
−ak + 2ak−1 + 6(k + 1) − 7 = −k 2 + 2(k − 1)2 + 6k + 6 − 7
−k 2 + 2(k 2 − 2k + 1) + 6k − 1 = −k 2 + 2k 2 − 4k + 2 + 6k − 1
k 2 + 2k + 1 = (k + 1)2 .
By the Strong Principle of Mathematical Induction, an = n2 for every positive integer n.