Sequences and Summations

This exercise set contains a lot of routine practice with the concept of and notation for sequences. It also discusses telescoping sums; the product notation, corresponding to the summation notation discussed in the section; and the factorial function, which occurs repeatedly in subsequent chapters. There are also a few challenging exercises on more complicated sequences and sums.

1. a) a 0=2·(-3)⁰+5°=2·1+1=3 b) a 1=2·(-3)¹+5¹=2·(-3)+5=-1

c) a4 = 2 · (-3)⁴ + 5⁴ = 2 · 81+625 = 787 d) a5 = 2 · (-3)⁵ + 5⁵ = 2 · (-243) + 3125 = 2639

3. In each case we simply evaluate the given function at n = 0, 1, 2, 3.

a) ao = 2° + 1 = 2, a 1 = 2¹ + 1 = 3, a2 = 2² + 1 = 5, a3 = 2³ + 1 = 9 b) ao = 1¹ = 1, ^a1 = 2² = 4, ^a2 = 3³ = 27, a3 = 4⁴ = 256

c) ao = L0/2J = 0, a1 = Ll/2J = 0, a2 = L2/2J = 1, a3 = L3/2j = 1

d) ao = L0/2j + I0/21 = 0 + 0 = 0, ai = Ll/2j + ll/21 = 0 + 1 = 1, az = L2/2j + 12/21 a3 = L3/2J + 13/21=1+2 = 3. Note that Ln/2J + ln/21 always equals n.

1+1 = 2,

5. In each case we just follow the instructions.

a) 2,5,8,11,14,17,20,23,26,29 b) 1,1,1,2,2,2,3,3,3,4 c) 1,1,3,3,5,5,7,7,9,9

d) This requires a bit of routine calculation. For example, the fifth term is 5! - 2⁵ = 120 - 32 = 88. The first ten terms are -1, -2, -2, 8, 88, 656, 4912, 40064, 362368, 3627776.

e) 3,6,12,24,48,96,192,384,768,1536

f)

2,4,6,10,16,26,42,68,110,178

g) For n = 1, the binary expansion is 1, which has one bit, so the first term of the sequence is 1. For n = 2, the binary expansion is 10, which has two bits, so the second term of the sequence is 2. Continuing in this way we see that the first ten terms are 1,2,2,3,3,3,3,4,4,4. Note that the sequence has one 1, two 2's, four 3's, eight 4's, as so on, with 2k-l copies of k.

h) The English word for 1 is "one" which has three letters, so the first term is 3. This makes a good brain-teaser; give someone the sequence and ask her or him to find the pattern. The first ten terms are 3,3,5,4,4,3,5,5,4,3.

7. One pattern is that each term is twice the preceding term. A formula for this would be that the nth term is 2n-l. Another pattern is that we obtain the next term by adding increasing values to the previous term.

Thus to move from the first term to the second we add 1; to move from the second to the third we add 2 ; then add 3, and so on. So the sequence would start out 1, 2, 4, 7, 11, 16, 22, .... We could also have trivial answers such as the rule that the first three terms are 1, 2, 4 and all the rest are 17 (so the sequence is 1, 2, 4, 17, 17, 17, ... ), or that the terms simply repeat 1, 2, 4, 1, 2, 4, 1, 2, 4, .... Here is another pattern: Take n points on the unit circle, and connect each of them to all the others by line segments. The inside of the circle will be divided into a number of regions. What is the largest this number can be? Call that value an.

If there is one point, then there are no lines and therefore just the one original region inside the circle; thus a1 = 1. If n = 2, then the one chord divides the interior into two parts, so a₂ = 2. Three points give us a triangle, and that makes four regions (the inside of the triangle and the three pieces outside the triangle), so a3 = 4. Careful drawing shows that the sequence starts out 1, 2, 4, 8, 16, 31. That's right: 31, not 32.

Section 2.4 Sequences and Summations 69

Creative students may well find other rules or patterns with various degrees of appeal.

9. We need to compute the terms of the sequence one at a time, since each term is dependent upon one or more of the previous terms.

a) We are given ao

=

2. Then by the recurrence relation an

=

6an-1 we see (by letting n

= 1)

that a1

=

6ao

=

6·2

=

12. Similarly a 2

=

_6a1

=

6·12

=

72, then a3

=

_6a2

=

6·72

=

432, and a4

=

_6a3

=

6·432

=

2592.

b) a1

=

2 (given), a2

= ai =

22 = 4,

a

3 =a~= 42 = 16, a4 =a~= 162 = 256, a5 =a~= 2562 = 65536 c) This time each term depends on the two previous terms. We are given a₀

=

1 and a₁

=

2. To compute a2 we let n

=

2 in the recurrence relation, obtaining a 2 = a1 + 3a0 = 2 + 3 · 1 = 5. Then we have a3 = a2 + 3a1 = 5 + 3 · 2

=

11 and a4 = a3 + 3a2 = 11 + 3 · 5 = 26.

d) ao = 1 (given), a1 = 1 (given), a 2 = 2a1+22ao = 2 · 1+4 · 1 = 6, a3 = 3a2 + 32a1 =3·6+9 · 1 = 27, a4 = 4a3 + 42a2 = 4 · 27 + 16 · 6 = 204

e) We are given ao

=

1, ai = 2, and a 2 = 0. Then a3 = a2 + a0 = 0 + 1 = 1 and a4 = a3 + a 1 = ¹ + 2 = 3.

11. a) We simply plug in n = 0, n = 1, n = 2, n = 3, and n = 4. Thus we have a₀ = 2° + 5 · 3° = 1 + 5 · 1 = 6, a1 = 21 + 5 · 3I = 2 + 5 · 3 = 17, a 2 = 22 + 5 · 32 = 4 + 5 · 9 = 49, a₃ = 2³ + 5 · 3³ = S + 5 · 27 = 143, and a4 = 2⁴ + 5 · 3⁴ = 16 + 5 · Sl = 421.

b) Using our data from part (a), we see that 49 = 5 · 17 - 6 · 6, 143 = 5 · 49 - 6 · 17, and 421 = 5 · 143 - 6 · 49.

c) This is algebra. The messiest part is factoring out a large power of 2 and a large power of 3. If we substitute n - 1 for n in the definition we have an-1 = 2n-l + 5 · 3n-I; similarly an_ 2 = 2n-2 + 5 · 3n-2. We start with the right-hand side of our desired identity:

5an-I - 6an-2 = 5(2n-l + 5 · 3n-l) - 6(2n- 2 + 5 · 3n- 2)

= 2n- 2(10 - 6) + 3n-2(75 - 30)

= 2n- 2 · 4 + 3n-2 · 9 · 5

= 2n

+

3n . 5 = an

13. In each case we have to substitute the given equation for an into the recurrence relation an =San-I -16an_ 2 and see if we get a true statement. Remember to make the appropriate substitutions for n (either n - 1 or n - 2) on the right-hand side. What we are really doing here is performing the inductive step in a proof by mathematical induction: if the formula is correct for an-I (and also for an- 2, etc., in some cases), then the formula is also correct for an .

a) Plugging an

=

0 into the equation an

=

8an-1 - 16an__{2 ,} we obtain the true statement that 0 = 0.

Therefore an

=

0 is a solution of the recurrence relation.

b) Plugging an = 1 into the equation an =San-I -16an-2, we obtain the false statement 1 = S· 1-16· 1 = -S.

Therefore an = 1 is not a solution.

c) Plugging an= 2n into the equation an= 8an-l -16an__{2 ,} we obtain the statement 2n = s.2n- 1-15.2n- 2.

By algebra, the right-hand side equals 2n-²(8 · 2 - 16) = 0. Since this is not equal to the left-hand side, we conclude that an

=

2n is not a solution.

d) Plugging an= 4n into the equation an= San-I -16an-2, we obtain the statement 4n = s.4n-l _ 15.4n-².

By algebra, the right-hand side equals 4n-²(S · 4-16) = 4n-² • 16 = 4n-² · 42 = 4n. Since this is the left-hand side, we conclude that an

=

4n is a solution.

e) Plugging an = n4n into the equation an= 8an-l -16an__{2 ,} we obtain the statement n4n = 8(n1)4nl -l6(n-2)4n-². By algebra, the right-hand side equals 4n-² (S(n-1)·4-16(n-2)) = 4n-²(32n-32-16n+32) = 4n- 2(l6n) = 4n-² · 4²n = n4n. Since this is the left-hand side, we conclude that an= n4n is a solution.

70 Chapter 2 Basic Structures: Sets, Functions, Sequences, Sums, and Matrices

f) Plugging a,, = 2 · 4n + 3n4" into the equation an = 8an-l - 16a,,_2, we obtain the statement 2 · 4n + 3n4"

=

8(2 · 4n-l + 3(n - 1)4n-l) - 16(2 · 4n- 2 + 3(n - 2)4n- 2). By algebra, the right-hand side equals 4"-²(8 · 2 ·4+8 ·3(n-1) · 4-16 · 2-16 · 3(n-2))

=

4n- 2(64+96n-96-32-48n+96)

=

4n-2(48n+32)

=

4n-² ·42(3n+2)

=

(2+3n)4n. Since this is the same as the left-hand side, we conclude that an= 2.4n +3n4n is a solution.

g) Plugging an= (-4)n into the equation a,, = 8an_1-l6a11_2, we obtain the statement (-4)" = 8·(-4)"-¹

-16· (-4)n-2. Byalgebratheright-handsideequals (-4)"- 2(8·(-4)-16)

=

(-4)n- 2(-48)

=

-3(-4)n. Since this is not equal to the left-hand side, we conclude that an= (-4)n is not a solution.

h) Plugging a,,

=

n²4n into the equation an

=

8a11_1 - l6a11 _2 , we obtain the statement n²4n

=

8(n -1)24n-¹-16(n-2)²4n- 2 . By algebra, the right-hand side equals 4n-² (8(n-1) 2 ·4-16(n-2)²) = 4n-2 (32(n2-2n + 1) -16(n2 - 4n+ 4))

=

4n- 2(32n² - 64n + 32 -16n2 + 64n- 64)

=

4"-²(16n² - 32)

=

4n-² · 4²(n² - 2)

=

4"(n² - 2). Since this is not equal to the left-hand side, we conclude that an= n 24n is not a solution.

15. In each case we have to plug the purported solution into the right-hand side of the recurrence relation and see if it simplifies to the left-hand side. The algebra can get tedious, and it is easy to make a mistake.

a) We have

an-l + 2an-2 + 2n - 9

=

-(n - 1) + 2 + 2(-(n - 2) + 2) + 2n - 9

= -n+2 =an.

b) We have

an-l + 2an-2 + 2n - 9 = 5(-1)"-^{1 -} (n - 1) + 2 + 2(5(-1)¹¹- 2 - (n - 2) + 2) + 2n - 9

= 5(-1)"-2(-1+2) - n + 2 =a,,.

Note that we had to factor out ( -1)^{11- 2} and that this is the same as ( -1 )" since ( -1 )²

=

1.

c) We have

an-l + 2a,,_2 + 2n - 9 = 3(-l)n-l + 2n-l - (n - 1) + 2 + 2(3(-1)"- 2 + 2n-2 - (n - 2) + 2) + 2n - 9

=

3(-1)"-2(-1+2) + 2n- 2(2 + 2) - n + 2 =a,,.

Note that we had to factor out 2^11- 2 and that 2n-² · 4

=

2".

d) We have

an-l + 2an-2 + 2n - 9 = 7 · 2n-l - (n - 1) + 2 + 2(7 · 2"- 2 - (n - 2) + 2) + 2n - 9

= 2n- 2(7 · 2 + 2 · 7) - n + 2 =an.

17. In the iterative approach, we write a,, in terms of a,,__{1 ,} then write an-l in terms of an-2 (using the recurrence relation with n - 1 plugged in for n), and so on. When we reach the end of this procedure, we use the given initial value of a_{0 .} This will give us an explicit formula for the answer or it will give us a finite series, which we then sum to obtain an explicit formula for the answer.

a) We write

an

=

3an-l

= 3(3a11-2) = 32an-2

= 32(3an-3) = 3³an-3

Note that we figured out the last line by following the pattern that had developed in the first few lines.

Therefore the answer is an = 2 · 3" .

Section 2.4 Sequences and Summations

b) We write

an= 2 +an-I

=

2 + (2 + an-2)

=

(2 + 2) + an-2

=

(2 · 2) + an-2

= (2 · 2) + (2 + an-3) = (3 · 2) + an-3

=

(n · 2) + an-n

=

(n · 2) + ao

=

(n · 2) + 3

=

2n + 3.

71

Again we figured out the last line by following the pattern that had developed in the first few lines. Therefore the answer is an = 2n

+

3.

c) We write (note that it is more convenient to put the an-I at the end) an= n +an-I

=

n + ((n - 1) + an-2)

=

(n + (n - 1)) + an-2

= (n + (n - 1)) + ((n - 2) + an-3) = (n + (n - 1) + (n - 2)) + an-3

= (n + (n - 1) + (n - 2) + · · · + (n - (n -1))) + an-n

=

(n + (n - 1) + (n - 2) + · · · + 1) + ao n(n + 1)

1_n²+n+2

2

+ -

2

Therefore the answer is an= (n²

+

n

+

2)/2. The formula used to obtain the last line-for the sum of the first n positive integers~is given in Table 2.

d) We write

an = 3

+

2n +an-I

=

3 + 2n + (3 + 2(n - 1) + an-2)

=

(2 · 3 + 2n + 2(n - 1)) + an-2

=

(2 · 3 + 2n + 2(n - 1)) + (3 + 2(n - 2) + an-3)

=

(3 · 3 + 2n + 2(n - 1) + 2(n - 2)) + an-3

=

(n · 3 + 2n + 2(n - 1) + 2(n - 2) + · · · + 2(n - (n - 1))) + an-n

=

(n · 3 + 2n + 2(n - 1) + 2(n - 2) + · · · + 2 · 1) + a0

=

3n + 2 · n(n

2+ l) + 4

=

n² + 4n + 4.

Therefore the answer is an

=

n²

+

4n

+

4. Again we used the formula for the sum of the first n positive integers from Table 2.

e) We write

an= -1+2an-I

= -1+2(-1+2an-2) = -3 + 4an-2

=

-3 + 4(-1+2an-3)

=

-7 + 8an-3

=

-7 + 8(-1+2an-4)

=

-15 + l6an-4

=

-15 + 16(-1+2an-5)

=

-31+32an-5

=

-(2n - 1) + 2nan-n

=

-2n + 1 + 2n · 1

=

1.

This time it was somewhat harder to figure out the pattern developing in the coefficients, but it became clear after we carried out the computation far enough. The answer, namely that an

=

1 for all n, it is clear in retrospect, after we found it, since 2 · 1 - 1

=

1.

72

f) We write

Chapter 2 Basic Structures: Sets, Functions, Sequences, Sums, and Matrices

an= 1+3an-l

= 1+3(1+3an-2) = (1 + 3) + 32an-2

= (1 + 3) + 32

(1 + 3an-3) = (1 + 3 + 32) + 3³an-3

=

(l+3+32 +···+3n-l)+3nan-n

= 1 + 3 + 32 + · · · + 3n-l + 3n 3n+l - 1

3-1 3n+l -1

2

(a geometric series)

Thus the answer is an

=

(3n+l - 1) /2.

g) We write

h) We write

an= nan-l = n(n - l)an-2

= n(n - l)(n - 2)an-3 = n(n - l)(n - 2)(n - 3)an-4

= n(n - l)(n - 2)(n - 3) · · · (n - (n - 1)) an-n

= n(n - l)(n - 2)(n - 3) · · · 1 · ao

= n! · 5 = 5n!.

= 2n(2(n - l)an-2) = 22

(n(n - l))an-2

=

22(n(n -1)) (2(n - 2)an-3)

=

23(n(n - l)(n - 2))an-3

= 2nn(n - l)(n - 2)(n - 3) · · · (n - (n - l))an-n

= 2nn(n - l)(n - 2)(n - 3) · · · 1 · ao

19. a) Since the number of bacteria triples every hour, the recurrence relation should say that the number of bacteria after n hours is 3 times the number of bacteria after n - 1 hours. Letting bn denote the number of bacteria after n hours, this statement translates into the recurrence relation bn = 3bn-l.

b) The given statement is the initial condition b₀

=

100 (the number of bacteria at the beginning is the number of bacteria after no hours have elapsed). We solve the recurrence relation by iteration: bn

=

3bn-l

=

32bn_ 2

=

· · · =

3nbn-n

=

3nbo. Letting n

=

10 and knowing that b0

=

100, we see that b10

=

3¹⁰ · 100

=

5,904,900.

21. a) Let ^Cn be the number of cars produced in the first n months. The initial condition could be taken to be c0

=

0 (no cars are made in the first 0 months). Since n cars are made in the nth month, and since ^Cn-l

cars are made in the first n - 1 months, we see that Cn = Cn-l

+

n.

b) The number of cars produced in the first year is c12 . To compute this we will solve the recurrence relation and initial condition, then plug in n = 12 (alternately, we could just compute the terms c1 , c2 , ... , c12

Section 2.4 Sequences and Summations

directly from the definition). We proceed by iteration exactly as we did in Exercise 17c:

Cn = n

+

^Cn-1

=

n

+

((n -1)

+

^Cn-2)

=

(n

+

(n - 1))

+

^Cn-2

=

(n

+

(n

-1)) +

((n -

2) +

^Cn-3)

=

(n

+

(n

-1) +

(n -

2)) +

Cn-3

= (n

+

(n - 1)

+

(n - 2)

+ · · · +

(n - (n - 1)))

+

c,,_n

=

(n

+

(n -

1) +

(n -

2) + · · · + 1) +co

n(n+l) n²+n

2 +O= 2

-Therefore the number of cars produced in the first year is (12²

+

12) /2 = 78.

c) We found the formula in our solution to part (b).

73

23. Each month our account accrues some interest that must be paid. Since the balance the previous month is B(k - 1), the amount of interest we owe is (0.07 /12)B(k - 1). After paying this interest, the rest of the $100 payment we make each month goes toward reducing the principal. Therefore we have B(k) =

B(k- l) - (100- (0.07/12)B(k-1)). This can be simplified to B(k)

=

(1

+

(0.07/12))B(k-1) -100. The initial condition is B(O)

=

5000. If one calculates this as k goes from 0 to 60, we see the balance gradually decrease and finally become negative when k

=

60 (i.e., after five years).

25. In some sense there are no right answers here. The solutions stated are the most appealing patterns that the author has found.

a) It looks as if we have one 1 and one 0, then two of each, then three of each, and so on, increasing the number of repetitions by one each time. Thus we need three more 1 's (and then four O's) to continue the sequence.

b) A pattern here is that the positive integers are listed in increasing order, with each even number repeated.

Thus the next three terms are 9, 10, 10.

c) The terms in the odd locations (first, third, fifth, etc.) are just the successive terms in the geometric sequence that starts with 1 and has ratio 2, and the terms in the even locations are all 0. The nth term is 0 if n is even and is 2<n-l)/² if n is odd. Thus the next three terms are 32, 0, 64.

d) The first term is 3 and each successive term is twice its predecessor. This is a geometric sequence. The nth term is 3 · 2n-l. Thus the next three terms are 384, 768, 1536.

e) The first term is 15 and each successive term is 7 less than its predecessor. This is an arithmetic sequence.

The nth term is 22 - 7n. Thus the next three terms are -34, -41, -48.

f) The rule is that the first term is 3 and the nth term is obtained by adding n to the (n - l)th term.

One can actually find a quadratic expression for a sequence in which the successive differences form an arithmetic sequence; here it is (n²

+

n

+

4)/2. The easiest way to see this is to note that the nth term is 3

+

2

+

3

+

4

+

5

+

6

+ · · · +

n. Except for the initial 3 instead of a 1 , the nth term is the sum of the first n positive integers, which is n(n

+

1)/2 by a formula in Table 2. Therefore the nth term is (n(n

+

1)/2)

+

2, as claimed. We see that the next three terms are 57, 68, 80.

g) One should play around with the sequence if nothing is apparent at first. Here we note that all the terms are even, so if we divide by 2 we obtain the sequence 1, 8, 27, 64, 125, 216, 343, .... This sequence appears in Table 1; it is the cubes. So the nth term is 2n3 . Thus the next three terms are 1024, 1458, 2000.

h) These terms look close to the terms of the sequence whose nth term is n! (see Table 1). In fact, we see that the nth term here is n!

+

1. Thus the next three terms are 362881, 3628801, 39916801.

74 Chapter 2 Basic Structures: Sets, Functions, Sequences, Sums, and Matrices

27. It is pretty clear that an should be approximately equal to n + yin, since the sequence is just the sequence of positive integers with perfect squares left out. There are about yin perfect squares up to n, so the count needs to get ahead by about this amount. Proving that this plausibility argument gives the correct formula involves some careful counting.

The sequence begins 2, 3, 5, 6, 7, 8, 10, 11, .... We can write it as the sequence an = n plus a sequence bn that jumps every time a perfect square is encountered. Thus {bn} begins 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, ... ; there are two l's, four 2's, six 3's, eight 4's, and so on. So we must show that bn = {yin}, where {yin}

means the integer closest to

fa

(note that there is never an ambiguity here, since this will never be a half-integer). Because of the way the sequence is formed, bn :::; k if and only if 2 + 4 + 6 + · · · + 2k 2". n. This is equivalent to k(J,~ + 1) 2:: n. Applying the quadratic formula and recalling that k is an integer, we obtain bn =

I (

-1 + vl +

4n)

/21 . Simplifying, we have bn =

I - ^~

⁺

Jn

⁺

~ I ·

Subtracting

~

and then rounding up is the same as rounding to the nearest integer (the smaller one if

F+"f

is a half-integer-see Exercise 47 in Section 2.3), so (with this understanding) bn = {

jn

⁺

^~}.

But it can never happen that yin:::; m +

~

^while

P ^>

^{m +}

^~

for some positive integer m-this would imply that n :::; m² + m +

~

^{and n}

>

m² + m, an

impossibility. Therefore {yin}=

{Jn+~},

and we are done.

An alternative solution is provided in the answer section of the text.

29. a) 2+3+4+5+6=20 b) 1-2+4-8+16=11 c) 3+3+···+3=10·3=30

d) This series "telescopes": each term cancels part of the term before it (see also Exercise 35). The sum is (2 - 1) + (4 - 2) + (8 - 4) + ... + (512 - 256) = -1+512 = 511.

31. We use the formula for the sum of a geometric progression: L~=O arJ = a(rn+I - l)/(r - 1).

a) Here a= 3, r = 2, and n = 8, so the sum is 3(2⁹ - 1)/(2 - 1) = 1533.

b) Here a = 1, r = 2, and n = 8. The sum taken over all the values of j from 0 to n is, by the formula, (2⁹ - 1)/(2 - 1) = 511. However, our sum starts at j = 1, so we must subtract out the term that isn't there, namely 2°. Hence the answer is 511 - 1 = 510.

c) Again we have to subtract the missing terms, so the sum is ((-3)⁹ - 1)/((-3) - 1) - (-3)⁰ - (-3)I = 4921 - 1 - (-3) = 4923.

d) 2((-3)⁹ - 1)/((-3) - 1) = 9842

33. The easiest way to do these sums, since the number of terms is reasonably small, is just to write out the summands explicitly. Note that the inside index (j) runs through all of its values for each value of the outside index ( i ).

a) (1+1) + (1+2) + (1+3) + (2 + 1) + (2 + 2) + (2 + 3) = 21 b) (0 + 3 + 6 + 9) + (2 + 5 + 8 + 11) + (4 + 7 + 10 + 13) = 78 c) (1+1+1) + (2 + 2 + 2) + (3 + 3 + 3) = 18

d) (0 + 0 + 0) + (1 + 2 + 3) + (2 + 4 + 6) = 18

35. If we just write out what the sum means, we see that parts of successive terms cancel, leaving only two terms:

n

L(a1 - a1-I) = aI - ao

+

a2 - aI

+

a3 - a2 +···+an-I - an-2 +an - an-I =an - ao

J=I

37. a) We use the hint, where ak = k²:

n n

k=I k=I

在文檔中 Student's Solutions Guide (頁 77-84)