Table Of ContentCHAPTER 1
1. (a) Substituting t for x2 yields t2 −14t+9 = 0. The quadratic formula
√ (cid:112) √
tells us that t = 7±2 10. Therefore x = ± 7±2 10.
√ √ √ √
(b) If x = 2 + 5, then x2 = 7 + 2 10 and x4 = 89 + 28 10. Thus
√ √ √
x4 − 14x2 + 9 = 0. If x = 2 − 5, then x2 = 7 − 2 10 and x4 =
√
89−28 10. Once again, x4 −14x2 +9 = 0. Since P(−x) = P(x), the facts
√ √ √ √ √ √ √ √
thatP( 2+ 5) = P( 2− 5) = 0,yieldP(− 2− 5) = P(− 2+ 5) = 0.
(cid:112) √ √ √ (cid:112) √ √ √
(c) 7+2 10 = 2+ 5 ≈ 3.65028, 7−2 10 = − 2+ 5 ≈ .82185,
(cid:112) √ √ √ (cid:112) √ √ √
− 7+2 10 = − 2− 5 ≈ −3.65028,− 7−2 10 = 2− 5 ≈ −.82185.
2. (a) Substituting t for x2 yields t2 −20t+16 = 0. The quadratic formula
√ (cid:112) √
tells us that t = 10±2 21. Therefore x = ± 10±2 21.
√ √ √ √
(b) If x = 3 + 7, then x2 = 10 + 2 21 and x4 = 184 + 40 21.
√ √ √
Thus x4 − 20x2 + 16 = 0. If x = 3 − 7, then x2 = 10 − 2 21 and
√
x4 = 184−40 21. Once again, x4 −20x2 +16 = 0. Since G(−x) = G(x),
√ √ √ √ √ √
the facts that G( 3 + 7) = G( 3 − 7) = 0, yield G(− 3 − 7) =
√ √
G(− 3+ 7) = 0.
(cid:112) √ √ √ (cid:112) √ √ √
(c) 10+2 21 = 3 + 7 ≈ 4.37780, 10−2 21 = − 3 + 7 ≈
(cid:112) √ √ √ (cid:112) √ √
.91370, − 10+2 10 = − 3 − 7 ≈ −4.37780, − 10−2 21 = 3 −
√
7 ≈ −.91370.
3. Substituting t for x2 yields t2−26t+81 = 0. The quadratic formula tells
√ (cid:112) √
us that t = 13±2 22. Therefore x = ± 13±2 22. Other techniques can
√ √
give the equivalent answers x = ± 2± 11.
4. Since (x + 1)5 = x5 + 5x4 + 10x3 + 10x2 + 5x + 1, (x + 1)5 + 32 =
x5 + 5x4 + 10x3 + 10x2 + 5x + 33 = 0. Therefore (x + 1)5 = −32, thus
x + 1 = −2 and x = −3. So −3 is the only real root. In the complex
numbers, all roots are of the form −1+ω, where ω5 = −32.
Also, (x+1)5 −21 = x5 +5x4 +10x3 +10x2 +5x−20 = 0. Therefore
(x+1)5 = 21, thus x+1 = 211 and x = −1+211. So −1+211 is the only
5 5 5
real root. In the complex numbers, all roots are of the form −1+ν, where
ν5 = 21.
5. Since (x − 2)5 = x5 − 10x4 + 40x3 − 80x2 + 80x − 32, (x − 2)5 − 29 =
x5 − 10x4 + 40x3 − 80x2 + 80x − 61 = 0. Therefore (x − 2)5 = 29, thus
1
2
x − 2 = 291 and x = 2 + 291. So 2 + 291 is the only real root. In the
5 5 5
complex numbers, all roots are of the form 2+ω, where ω5 = 29.
6. Since (x+3)5 = x5+15x4+90x3+270x2+405x+243, (x+3)5+117 =
x5+15x4+90x3+270x2+405x+360 = 0. Therefore (x+3)5 = −117, thus
x+3 = −1171 and x = −3−1171. So −3−1171 is the only real root. In
5 5 5
the complex numbers, all roots are of the form −3+ω, where ω5 = −117.
7. Substituting t for x3 yields t2 + 10t + 21 = 0. Since t2 + 10t + 21 =
(t+3)(t+7), we have t = −3 or t = −7. Therefore x = −31,−71 are the
3 3
only real roots. In the complex numbers, all roots are of the form −α or −β,
where α3 = 3 and β3 = 7.
8. If (x2 −1)4 −23 = 0, then (x2 −1)4 = 23 and x2 −1 = ±231. Therefore
(cid:112) 4
x2 = 1±231. Since 1−231 < 0, the only real roots are ± 1+231. In the
4 4 √ 4
complex numbers, all roots are of the form ± 1+ω, where ω4 = 23.
9. 15◦, 75◦, and 105◦ can be constructed as 15◦ = 45◦−30◦, 75◦ = 30◦+45◦,
and 105◦ = 75◦+30◦. Since 20◦ cannot be constructed, neither can 10◦, 40◦,
95◦. Observe that if any of these angles could be constructed, then we could
construct 20◦ as 20◦ = 10◦ +10◦, 20◦ = 40◦, 20◦ = 95◦ −(30◦ +45◦).
2
10. 3◦, 6◦, 9◦ can be constructed as 3◦ = 36◦−30◦, 6◦ = 3◦+3◦, 9◦ = 3◦+6◦.
2
1◦, 2◦, 4◦, 5◦, 7◦, 8◦ cannot be constructed for if you could construct 1◦, 2◦,
4◦, 5◦, or 8◦, then you could construct 40◦, which is impossible. Finally, 7◦
cannot be constructed for if you could construct 7◦, then you could construct
2◦ = 30◦ −(7◦ +7◦ +7◦ +7◦), which is also impossible.
11. n◦ can be constructed if and only if n is a multiple of 3. Note that since
we can construct 3◦, we can construct any multiple of 3◦. Conversely, if n
is not a multiple of 3, then either 1◦ or 2◦ could be written as n◦ minus
a multiple of 3◦ which would lead to the contradiction that either 1◦ or 2◦
could be constructed.
12. (a) cos(75◦) = cos(30◦ + 45◦) = cos(30◦)cos(45◦) − sin(30◦)sin(45◦)
√ √ √ √ √
= 3 · 2 − 1 · 2 = 6− 2.
2 2 2 2 4
√ √ √ √
(b) If x = 6− 2, then x2 = 2− 3 and x4 = 7−4 3. Thus 16x4 −16x2 +
4 4 16
1 = 0.
√ √ √ √ √ √
(c) 6+ 2, − 6− 2− 6+ 2 are the other three roots.
4 4 4
√ √ √ √ √ √
(d) cos(15◦) = 6+ 2, cos(165◦) = − 6− 2, cos(105◦) = − 6+ 2.
4 4 4
3
13. Since cos2(θ)+αcos(θ)+β = 0, we have cos2(θ)+β = −αcos(θ). Squar-
ing both sides yields (cos2(θ))2 + 2βcos2(θ) + β2 = α2cos2(θ). Replacing
cos2(θ) by 1−sin2(θ) gives us (1−sin2(θ))2+2β(1−sin2(θ))+β2 = α2(1−
sin2(θ)). Thus sin4(θ)+(−2−2β)sin2(θ)+(β2+2β+1) = α2−α2sin2(θ),
which is equivalent to sin4(θ)+(α2−2−2β)sin2(θ)+(β2+2β+1−α2) = 0.
Thus sin(θ) is a root of x4 +(α2 −2β −2)x2 +(β2 +2β +1−α2).
14. The values of f (x) = f(x+5)−f(x) are all 15, therefore this table can
1
be produced by a linear function. f(x) = 3x+18 produces this table.
15. (a) The values of g (x) = g(x+2)−g(x) are not constant but the values
1
of g (x) = g (x+2)−g (x) are all 24. Therefore this table can be produced
2 1 1
by a quadratic function but not by a linear function.
(b) g(x) = 3x2−5x−11 produces this table. To findg(x), let g(x) = ax2+
bx+c and plug in three different values of x to obtain three linear equations
involving a,b,c. For example, using x = −1,1,3 yields the equations a−b+
c = −3, a+b+c = −13, 9a+3b+c = 1.
16. The values of h (x) = h(x+4)−h(x) are not constant but the values of
1
h (x) = h (x+4)−h (x) are all −32, therefore this table can be produced by
2 1 1
a quadratic function but not by a linear function. Plugging in three values
of x into h(x) = ax2+bx+c gives us three linear equations in a,b,c allowing
us to determine that h(x) = −x2 +10x+17.
17. If we let k (x) = k(x + 1) − k(x) and k (x) = k (x + 1) − k (x),
1 n+1 n n
for n ≥ 1, then the values of k (x),k (x),k (x),k (x) are not fixed but the
1 2 3 4
values of k (x) are all 240. Therefore the smallest degree polynomial which
5
canproducethistablehasdegree5. Ifyoutrytofindthedegree5polynomial
which produces this table, it is k(x) = 2x5 −4x4 +3x3 −6x2 +11x−9.
18. Multiplying both sides by x2+11x+28 and then equating the coefficients
ofx1 andx0 frombothsides,weobtaintheequationsa+b = 2,4a+7b = −19.
As a result, a = 11,b = −9.
19. Multiplying both sides by x3 + x and then equating the coefficients of
x2,x1,x0 from both sides, we obtain the equations a+b = 11, c = −2, a = 4.
As a result, a = 4,b = 7,c = −2.
20. There are many integer solutions to the equation 5A + 4B = 43. One
solution is A = 7,B = 2 and another is A = 3,B = 7.
21. (a) F(cid:48)(x) = 36x2 −16x−17 and its roots are −1, 17.
2 18
4
(b) −1 is also a root of F(x), therefore it is a double root of F(x). As a
2
result, 2x+1 is a factor of F(cid:48)(x) and (2x+1)2 is a root of F(x).
(c) F(x) = (2x+1)2(3x−5) and F(cid:48)(x) = (2x+1)(18x−17).
22. (a) G(cid:48)(x) = 5x4 −8 and its only real roots are ±(8)1.
4
5
(b) By plugging each of them into G(x), we see that neither ±(8)1 is a
4
5
root of G(x).
(c) G(x) and G(cid:48)(x) have no real roots in common, therefore G(x) not only
has no multiple roots in the real numbers but has no multiple roots in the
complex numbers.
5
CHAPTER 2
Sections 2.1, 2.2, 2.3
1. We will apply Mathematical Induction. The inequality is certainly true
for n = 1. Next, if we assume the inequality holds for some k ∈ N, then
(1+r)k+1 = (1+r)k(1+r) ≥ (1+rk)(1+r) = 1+r(k+1)+r2k ≥ 1+r(k+1)
and the inequality also holds for k +1.
2. We will apply Mathematical Induction. Both (a) and (b) are certainly
true for n = 1. Next, if we assume that (a) and (b) hold for some k ∈ N, then
ln(r ·r ···r ·r ) = ln((r ·r ···r )·r ) = ln(r ·r ···r )+ln(r ) =
1 2 k k+1 1 2 k k+1 1 2 k k+1
(ln(r )+ln(r )+···+ln(r ))+ln(r ) = ln(r )+ln(r )+···+ln(r )+ln(r )
1 2 k k+1 1 2 k k+1
and e(r1+r2+···+rk+rk+1) = e((r1+r2+···+rk)+rk+1) = e(r1+r2+···+rk) · erk+1 =
(er1 · er2 ···erk) · erk+1 = er1 · er2 ···erk · erk+1. Therefore (a) and (b) also
hold for k +1.
3. We will apply Mathematical Induction. Both (a) and (b) are certainly
true for n = 1. Next, if we assume that (a) and (b) hold for some k ∈ N, then
(xk+1)(cid:48) = (xk·x)(cid:48) = (xk)(cid:48)(x)+(xk)(x)(cid:48) = (kxk−1)(x)+(xk)(1) = kxk+xk =
(k+1)xk and(f(x)k+1)(cid:48) = (f(x)k·f(x))(cid:48) = (f(x)k)(cid:48)(f(x))+(f(x)k)(f(x))(cid:48) =
(kf(x)k−1f(cid:48)(x))(f(x))+(f(x)k)(f(cid:48)(x)) = kf(x)kf(cid:48)(x)+f(x)kf(cid:48)(x) = (k +
1)f(x)kf(cid:48)(x). Therefore (a) and (b) also hold for k +1.
4. We will apply Mathematical Induction. Both (a) and (b) are certainly
true for n = 1. Next, if we assume that (a) and (b) hold for some k ∈ N,
then F(r +r +···+r +r ) = F((r +r +···+r )+r ) = F(r +
1 2 k k+1 1 2 k k+1 1
r +···+r )+F(r ) = (F(r )+F(r )+···+F(r ))+F(r ) = F(r )+
2 k k+1 1 2 k k+1 1
F(r )+···+F(r )+F(r ) and G(r ·r ···r ·r ) = G((r ·r ···r )·
2 k k+1 1 2 k k+1 1 2 k
r ) = G(r · r ···r ) · G(r ) = (G(r ) · G(r )···G(r )) · G(r ) =
k+1 1 2 k k+1 1 2 k k+1
G(r )·G(r )···G(r )·G(r ). Therefore (a) and (b) also hold for k +1.
1 2 k k+1
5. WewillapplyMathematicalInduction. Equalitycertainlyholdsforn = 1.
Next, if we assume equality holds for some k ∈ N, then 2+5+8+···+(3(k+
1)−1) = 2+5+8+···+(3k−1)+(3k+2) = (2+5+8+···+(3k−1))+(3k+2) =
3k2+k+(3k+2) = 3k2+7k+4 = 3(k+1)2+(k+1) and equality also holds for k+1.
2 2 2
6. WewillapplyMathematicalInduction. Equalitycertainlyholdsforn = 1.
Next, if we assume equality holds for some k ∈ N, then 12 +32 +52 +···+
(2(k+1)−1)2 = 12+32+52+···+(2k−1)2+(2k+1)2 = (12+32+52+···+(2k−
1)2)+(2k+1)2 = 4k3−k +(4k2+4k+1) = 4k3+12k2+11k+3 = 4(k+1)3−(k+1)
3 3 3
6
and equality also holds for k +1.
7. WewillapplyMathematicalInduction. Equalitycertainlyholdsforn = 1.
Next, if we assume equality holds for some k ∈ N, then 13+23+33+···+(k+
1)3 = 13+23+33+···+k3+(k+1)3 = (13+23+33+···+k3)+(k+1)3 =
k2(k+1)2+(k+1)3 = (k+1)2(k2+(k+1)) = (k+1)2(k2+4k+4) = (k+1)2(k+2)2
4 4 4 4
and equality also holds for k +1.
8. We will apply Mathematical Induction. Since F = F = 1, the result
1 2
certainly holds for n = 1. Next, let us assume that the result holds for some
k ∈ N, therefore the largest integer that divides both F and F is one.
k k+1
Now suppose that c is a positive integer that divides both F and F .
k+1 k+2
Then c divides F − F = F . Thus c divides both F and F and
k+2 k+1 k k k+1
therefore must be equal to 1. Hence, the result also holds for k +1.
9. We will apply Mathematical Induction. Since F = F = 1 and F = 2,
1 2 3
we see that the result holds for n = 1. Next, let us assume that the result
holds for some k ∈ N, therefore |(F )2−F F | = |(F )(F +F )−
k+2 k+3 k+1 k+2 k k+1
(F +F )F | = |F F −(F )2| = |(F )2−F F | = 1 and the
k+1 k+2 k+1 k+2 k k+1 k+1 k+2 k
result also holds for k +1.
10. We will apply Mathematical Induction. Since r2−1 = r + 1, we see
r−1
that a+ar = ar2−a, thus equality holds for n = 1. If we assume the result
r−1
holds for some k ∈ N, we have a+ar +···+ark +ark+1 = (a+ar +···+
ark−1 +ark)+ark+1 = ark+1−a +ark+1 = (ark+1−a)+ark+1(r−1) = ark+2−a.
r−1 r−1 r−1
Therefore equality also holds for k +1.
11. We will apply Mathematical Induction. The inequality certainly holds
when n = 1. If we assume the inequality holds for some k ∈ N, then |r +r +
1 2
···+r +r | = |(r +r +···+r )+r | ≤ |r +r +···+r |+|r | ≤
k k+1 1 2 k k+1 1 2 k k+1
(|r |+|r |+···+|r |)+|r | = |r |+|r |+···+|r |+|r |. Therefore
1 2 k k+1 1 2 k k+1
the inequality also holds for k +1.
12. We will apply Mathematical Induction. When n = 1, both sides of the
equality are equal to 1. If we assume the inequality holds for some k ∈ N,
2
then 1 + 1 + 1 +···+ 1 = 1 + 1 + 1 +···+ 1 + 1 =
1·2 2·3 3·4 (k+1)(k+2) 1·2 2·3 3·4 k(k+1) (k+1)(k+2)
( 1 + 1 + 1 +···+ 1 )+ 1 = k + 1 = k2+2k+1 =
1·2 2·3 3·4 k(k+1) (k+1)(k+2) k+1 (k+1)(k+2) (k+1)(k+2)
(k+1)2 = k+1 . Therefore equality also holds for k +1.
(k+1)(k+2) (k+2)
13. We will apply Mathematical Induction. When n = 1, both sides of the
equality are equal to 1. If we assume the inequality holds for some k ∈ N,
3
7
then 1 + 1 + 1 +···+ 1 = 1 + 1 + 1 +···+ 1 + 1 =
1·3 2·4 3·5 (k+1)(k+3) 1·3 2·4 3·5 k(k+2) (k+1)(k+3)
( 1 + 1 + 1 + ··· + 1 ) + 1 = 3k2+5k + 1 =
1·3 2·4 3·5 k(k+2) (k+1)(k+3) 4(k+1)(k+2) (k+1)(k+3)
3k3+14k2+19k+8 = 3(k+1)3+5(k+1)2 = 3(k+1)2+5(k+1). Therefore equality
4(k+1)(k+2)(k+3) 4(k+1)(k+2)(k+3) 4(k+2)(k+3)
also holds for k +1.
14. We will apply Mathematical Induction. When n = 1, both sides of the
equality are equal to 1. If we assume the inequality holds for some k ∈ N,
4
then 1 + 1 + 1 +···+ 1 = 1 + 1 + 1 +···+ 1 + 1 =
1·4 2·5 3·6 (k+1)(k+4) 1·4 2·5 3·6 k(k+3) (k+1)(k+4)
( 1 + 1 + 1 +···+ 1 )+ 1 = 11k3+48k2+49k + 1 =
1·4 2·5 3·6 k(k+3) (k+1)(k+4) 18(k+1)(k+2)(k+3) (k+1)(k+4)
11k4+92k3+259k2+286k+108 = 11(k+1)4+48(k+1)3+49(k+1)2 =
18(k+1)(k+2)(k+3)(k+4) 18(k+1)(k+2)(k+3)(k+4)
11(k+1)3+48(k+1)2+49(k+1). Therefore equality also holds for k +1.
18(k+2)(k+3)(k+4)
15. We will apply Mathematical Induction. If n = 2, then there is only one
subset with 2 elements and the formula is correct. Now assume the formula
holds for some k ∈ N and suppose we are given a set with k + 1 elements.
Then it can be considered to be the union of a set S with k elements and an
additional element x. Therefore every subset of S ∪{x} with two elements
is a subset of S with 2 elements or the union of {x} with a subset of S
containing 1 element. Since S has k(k−1) subsets with 2 elements and k
2
subsets with 1 element, we see that S ∪{x} has k(k−1) +k = (k+1)k subsets
2 2
with 2 elements. Therefore the formula also holds for k +1.
16. We will apply Mathematical Induction. If n = 3, then there is only one
subset with 3 elements and the formula is correct. Now assume the formula
holds for some k ∈ N and suppose we are given a set with k + 1 elements.
Then it can be considered to be the union of a set S with k elements and an
additional element x. Therefore every subset of S∪{x} with three elements
is a subset of S with 3 elements or the union of {x} with a subset of S
containing 2 elements. Since S has k(k−1)(k−2) subsets with 3 elements and
6
k(k−1) subsets with 2 element (see exercise 15), we see that S ∪ {x} has
2
k(k−1)(k−2) + k(k−1) = (k+1)k(k−1) subsets with 3 elements. Therefore the
6 2 6
formula also holds for k +1.
17. We will apply Mathematical Induction. If n = 1, then the empty set
and the entire set are the only two subsets and the formula is correct. Now
assume the formula holds for some k ∈ N and suppose we are given a set
with k +1 elements. Then it can be considered to be the union of a set S
with k elements and an additional element x. Therefore we can partition the
8
subsets of S ∪{x} into two collections - those that contain x and those that
don’t. There are 2k subsets which do not contain x as these are the subsets
of S. However, the subsets which contain x are all of the form A ∪ {x},
where A is a subset of S. Therefore there are also 2k subsets which contain
x, hence the total number of subsets is 2k+2k = 2(2k) = 2k+1 subsets. Thus
the formula also holds for k +1.
18. We will apply Mathematical Induction to prove (a) and then use (a) to
prove (b). Since s = 11, (a) holds for n = 1. Next, assume that (a) holds
1
for some k ∈ N. Then s = 2(s +5) > 2(10+5) = 10, hence (a) holds for
k+1 3 k 3
k+1. Since (a) holds, we now have s −s = 2(s +5)−s = 10−sk < 0.
k+1 k 3 k k 3
Hence s < s and (b) also holds.
k+1 k
19. We will apply Mathematical Induction to prove (a) and then use (a) to
prove (b). Since t = 9, (a) holds for n = 1. Next, assume that (a) holds for
1
some k ∈ N. Then t = 2(t +5) < 2(10+5) = 10, hence (a) holds for
k+1 3 k 3
k+1. Since (a) holds, we now have t −t = 2(t +5)−t = 10−tk > 0.
k+1 k 3 k k 3
Hence t > t and (b) also holds.
k+1 k
20. We will apply Mathematical Induction to prove (a) and then use (a) to
prove (b). Since u = 5, (a) holds for n = 1. Next, assume that (a) holds for
1
some k ∈ N. Then u = 3(u +12) < 3(18+12) = 18, hence (a) holds for
k+1 5 k 5
k+1. Since(a)holds, wenowhaveu −u = 3(u +12)−u = 36−2uk > 0.
k+1 k 5 k k 5
Hence u > u and (b) also holds.
k+1 k
21. We will apply Mathematical Induction to prove (a) and then use (a) to
prove (b). Since v = 20, (a) holds for n = 1. Next, assume that (a) holds for
1
some k ∈ N. Then v = 3(v +12) > 3(18+12) = 18, hence (a) holds for
k+1 5 k 5
k+1. Since (a) holds, we now have v −v = 3(v +12)−v = 36−2vk < 0.
k+1 k 5 k k 5
Hence v < v and (b) also holds.
k+1 k
22. We will apply Mathematical Induction. The result it certainly true for
n = 1. Next, assume the result is true for some k ∈ N, then f (x)+f (x)+
1 2
···+f (x) = (f (x)+f (x)+···+f (x))+f . Since f (x)+f (x)+
k+1 1 2 k k+1 1 2
···+f (x) and f are two continuous functions, so is their sum. Therefore
k k+1
the result is also true for k +1.
23. We will apply Mathematical Induction. The result is certainly true for
n = 1. Next, assume the result is true for some k ∈ N, then F(k + 1) =
(k+1)F(k) = (k+1)(5(3kk−!1)) = 5((k+1)!). Thus the result is also true for k+1.
3 3 3k
24. We will apply Mathematical Induction. The result is certainly true for
9
n = 1, Next, assume the result is true for some k ∈ N, then G(k + 1) =
G(k)+3k2+7k+3 = (k3+2k2+5)+(3k2+7k+3) = k3+5k2+7k+8 =
(k3+3k2+3k+1)+2(k2+2k+1)+5 = (k+1)3+2(k+1)2+5. Thus the
result is also true for k +1.
25. We will apply Mathematical Induction. Since 62 > 6·6−5, the inequality
holds for n = 6. Next, assume the inequality holds for some integer k ≥ 6.
Then(k+1)2 = k2+2k+1 > (6k−5)+(2k+1) = 8k−4 > 6k+1 = 6(k+1)−5
and the inequality also holds for k +1.
26. We will apply Mathematical Induction. Since 92 > 11 · 9 − 24, the
inequality holds for n = 9. Next, assume the inequality holds for some
integer k ≥ 9. Then (k + 1)2 = k2 + 2k + 1 > (11k − 24) + (2k + 1) =
13k−23 > 11k−13 = 11(k+1)−24 and the inequality also holds for k+1.
27. We will apply Mathematical Induction. Since 1 + 1 < 2 − 1, the
12 22 2
inequality holds for n = 2. Next, assume the inequality holds for some
integer k ≥ 2. Observe that 1 < 1 = 1 − 1 which implies that
(k+1)2 k(k+1) k k+1
−1 + 1 < − 1 . We now have 1 + 1 + ··· + 1 = 1 + 1 +
k (k+1)2 k+1 12 22 (k+1)2 12 22
··· + 1 + 1 = ( 1 + 1 + ··· + 1 ) + 1 < (2 − 1) + 1 =
k2 (k+1)2 12 22 k2 (k+1)2 k (k+1)2
2+(−1 + 1 ) < 2− 1 . Therefore the inequality also holds for k +1.
k (k+1)2 k+1
28. We will apply Mathematical Induction. Since 13 +5·1 is a multiple of
3, the result holds for n = 1. Next, assume the result holds for some k ∈ N,
then (k +1)3 +5(k +1) = k3 +3k2 +8k +6 = (k3 +5k)+(3k2 +3k +6),
which is the sum of two multiples of 3. Therefore the result also holds for
k +1.
29. Let S be the elements of N which do not belong to T. Our goal is to
prove that S is the empty set. If S is not empty, the Well Ordering Principle
guarantees that S has a smallest element s. Since 1 ∈ T, we know that s > 1.
Therefore T contains all elements of the set {1,...,s−1}. Thus T contains
s = (s−1)+1, which is a contradiction. Hence S is empty and T = N.
30. Let V = {t − m + 1 | t ∈ T and t ≥ m}. Then V is a subset of N
which satisfies (a) and (b) of the first version of Mathematical Induction.
Therefore V = N. As a result, if s is an integer greater than or equal to m,
then s−m+1 > 0, thus s−m+1 ∈ V. Thus there exist some t ∈ T such
that s−m+1 = t−m+1, hence s = t ∈ T, proving the result.
31. Let S be the set of positive even integers which do not belong to T. Our
goal is to prove the S is the empty set. If S is not empty, the Well Ordering
10
Principle guarantees that S has a smallest element s. Since 2 ∈ T, we know
that s > 2. Thus T contains s−2, hence T contains s = (s−2)+2, which
is a contradiction. Hence T contains all positive even integers.
32. Let S be the set of positive integers which are multiples of three and do
not belong to T. Our goal is to prove the S is the empty set. If S is not
empty, the Well Ordering Principle guarantees that S has a smallest element
s. Since 3 ∈ T, we know that s > 3. Thus T contains s−3, hence T contains
s = (s−3)+3, which is a contradiction. Hence T contains all positive even
integers.
33. (a) We will apply Mathematical Induction. Since 91−1 is a multiple of 8,
the result is true for n = 1. Next, assume there is some k ∈ N such that the
result is true, then 9k+1−1 = (9k+1−9k)+(9k−1) = 9k·(9−1)+(9k−1) =
9k·8+(9k−1), which is the sum of two multiples of 8. Therefore the result
also holds for k +1.
(b) We will use the version of Mathematical Induction from exercise 31.
Since 92 −1 is a multiple of 80, the result is true for n = 2. Next, assume
there is some k ∈ N such that the result is true, then 9k+2 − 1 = (9k+2 −
9k)+(9k−1) = 9k·(92−1)+(9k−1) = 9k·80+(9k−1), which is the sum
of two multiples of 80. Therefore the result also holds for k +2.
34. (a) We will use the version of Mathematical Induction from exercise
31. Since 72 − 1 is a multiple of 48, the result is true for n = 2. Next,
assume there is some k ∈ N such that the result is true, then 7k+2 − 1 =
(7k+2−7k)+(7k −1) = 7k ·(72−1)+(7k −1) = 7k ·48+(7k −1), which is
the sum of two multiples of 48. Therefore the result also holds for k +2.
(b) We will use a version of Mathematical Induction which applies to the
set of positive integers which are multiples of 4. Since 74 − 1 is a multiple
of 2400, the result is true for n = 4. Next, assume there is some k ∈ N
such that the result is true, then 7k+4 − 1 = (7k+4 − 7k) + (7k − 1) =
7k·(74−1)+(7k−1) = 7k·2400+(7k−1), which is the sum of two multiples
of 2400. Therefore the result also holds for k +4.
35. (a) Letting n = 1,2,3, yields the equations a+b+c = 1,4a+2b+c =
4,9a+3b+c = 9.
(b) a = 1,b = 0,c = 0 yielding the quadratic polynomial n2.
(c) We already know that the formula works for n = 1,2,3. Next, assume
the formula works for some k ∈ N, then 1+3+5+···+(2(k +1)−1) =
