Solutions for practice questions: Chapter 3, Sequences and Series If you find any errors, please let me know at mailto:peggy.frisbie@polk-fl.net. First of all, there are more than 60 problems in this set! I am starting to wonder about my decision to work all the problems. For now, though, I’m still on it. Here we go. 1. s(n) = 2n – 3, find the first five terms. No problem. s(1) = 2 · 1 – 3 = 2 – 3 = –1 s(2) = 2 · 2 – 3 = 4 – 3 = 1 s(3) = 2 · 3 – 3 = 6 – 3 = 3 s(4) = 2 · 4 – 3 = 8 – 3 = 5 s(5) = 2 · 5 – 3 = 10 – 3 = 7 2. g(k) = 2k – 3 Seriously, g and k? g(1) = 21 – 3 = 2 – 3 = –1 g(2) = 22 – 3 = 4 – 3 = 1 g(3) = 23 – 3 = 8 – 3 = 5 g(4) = 24 – 3 = 16 – 3 = 13 g(5) = 25 – 3 = 32 – 3 = 29 3. f (n) = 3 ´ 2–n

f (1) = 3 ´ 2–1 =

f (2) = 3 ´ 2–2 =

f (3) = 3 ´ 2–3 =

f (4) = 3 ´ 2–4 =

f (5) = 3 ´ 2–5 =

4.

5. This one is a little weird. The variable that’s

changing is n. It will be 1, 2, 3, 4, and 5. The k, on the other hand, is not?

And then I checked the back of the book. Apparently they intended for the k to be n, too. That’s BoB’s mistake, not mine, and I’m not changing my answers. If you were to let k = 1, 2, 3, 4, and 5 as you go down the list, you should get his answers.

6.

7. If I were writing this set of questions, I

might have had fewer of each type of problem. But I’m just working them. So, is 52, 55, 58, 61, … arithmetic, geometric, or neither? Given that the values increase by 3 units between successive terms, I’m going with arithmetic. And that number, 3, is called the common difference.

8. –1, 3, –9, 27, –81, … is geometric. I can tell

because , and they’re

all equal to –3, which is therefore the common ratio.

323438316332

1

1

53; for 1n n

aa a n-

=ìí = + >î1

2

3

4

5

55 3 88 3 1111 3 1414 3 17

aaaaa

== + == + =

= + == + =

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

( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )

11

22

33

44

55

1 2 3 1 2 3 3 2

1 2 3 1 2 3 3 2

1 2 3 1 2 3 3 2

1 2 3 1 2 3 3 2

1 2 3 1 2 3 3 2

k k k

k k k

k k k

k k k

k k k

a

a

a

a

a

= - + = - × + = -

= - + = + × + = +

= - + = - × + = -

= - + = + × + = +

= - + = - × + = -

1

1

32 ; for 2n n

bb b n n-

=ìí = + ³î

( )( )( )( )

1

2

3

4

5

33 2 2 7

7 2 3 13

13 2 4 21

21 2 5 31

bb

b

b

b

=

= + =

= + =

= + =

= + =

3 9 27 811 3 9 27

- -= = =

- -

9. 0.1, 0.2, 0.4, 0.8, 1.6, 3.2, … is geometric. Each time, the value is doubled, so the common ratio is 2.

10. 3, 6, 12, 18, 21, 27, … is finally more

interesting! Initially, it certainly looked to me like the terms were doubling; 3 · 2 = 6, and 6 · 2 = 12. But then 12 · 2 ≠ 18. And then I wondered if I had read the first term incorrectly, and maybe it was adding 6, as 6 + 6 = 12 and 12 + 6 = 18. But that definitely fails, too. Not only is 3 + 6 ≠ 6, but 18 + 6 ≠ 21. So the sequence is neither arithmetic nor geometric. But that raises another question1 — is it following some predictable pattern?

Examining the differences between successive terms gives +3, +6, +6, +3, +6. So I suppose that the pattern could be add three, then add six twice, then start over again. Without more evidence, it’s hard to know for sure.

11. 6, 14, 20, 28, 34, … is once again neither

arithmetic nor geometric, but this time the pattern seems easier to detect. The differences are +8, +6, +8, +6. I’m thinking that it would just alternate adding 8 and 6 as it continues.

12. 2.4, 3.7, 5, 6.3, 7.6, … is arithmetic again.

The difference between each pair of terms is 1.3.

13. Yeah, I’d have had fewer of each type. It

makes me wonder if Mr. Wazir and Mr. Garry split up the task of writing the questions by chapter, and one of them likes a lot more practice examples than the other.

–3, 2, 7, 12, … is pretty clearly arithmetic, with a common difference of 5.

1 Note: It’s “raises a question,” not “begs a question.” The phrase “to beg the question” is about sidestepping an issue, not actively bringing it up! Lots and lots and lots of people use it incorrectly. I am sometimes tempted to say it incorrectly myself. But as I hope to avoid sounding uneducated to those who are well-educated, I try really hard to get it right.

a) To find the eighth term, I’ll add 5 a few more times: 17, 22, 27, 32. That’s it: 32.

b) The general formula for the nth term of an arithmetic sequence is

. In this case, = –3, and d = 5. The nth term is

thus = –3 + 5n – 5 = 5n – 8. And yes, I could have found this before I did part (a), and then used it to get the answer to part (a), but that seemed to subvert the structure of the question.

c) A recursive definition for a seque22nce has two (at least) parts. The first is an initial term (or terms), and the second is a rule for how to proceed from there. In this case, the first term is –3, and the rule for proceeding is “add 5 to the previous term.” In symbols, that looks like this:

.

Having looked at the answer in the back now, I know that your book uses a for the dependent variable rather than u. Either is okay, but I’m sticking with u, because in my experience, the exams are more likely to use u.

14. Fewer words this time. 19, 15, 11, 7, … is

arithmetic, with first term 19 and common difference –4. a) Continuing to add –4 gives 3, –1, –5,

–9. b) = 19 + (n – 1)(–4) =

19 – 4n + 4 = 23 – 4n.

c)

15. –8, 3, 14, 25, … is arithmetic, with first

term –8 and common difference 11. This is getting tedious. a) 36, 47, 58, 69. Hmmm. No subtext

there.

( )1 1nu u n d= + -

1u

( )3 1 5nu n= - + - ×

1

1

35, 1n n

uu u n-

= -ìí = + >î

( )1 1nu u n d= + -

1

1

194, 1n n

uu u n-

=ìí = - >î

b) = –8 + (n – 1) · 11 = –8 + 11n – 11 = 11n – 19.

c)

Whoa! Did you see how I mixed it up there? Keeping it interesting!

16. 10.05, 9.95, 9.85, 9.75, … Once again, it’s looking arithmetic, this

time with a common difference of –0.1. a) 9.65, 9.55, 9.45, 9.35. b) = 10.05 + (n – 1)(–0.1) = 10.05 – 0.1n + 0.1 = 10.15 – 0.1n

c)

17. 100, 99, 98, 97, … is once again

arithmetic2, with first term 100 and common difference –1. a) 96, 95, 94, 93…3 b) = 100 + (n – 1) · –1 = 100 – n + 1 = 101 – n

c)

18. is arithmetic, with a

common difference of .

a) . I find this one

much easier to think about in terms of subtracting the fractions if I just keep the common denominator going.

b) =

.

2 Hey, look, another arithmetic sequence. Please just shoot me now. 3 And the anesthesia has kicked in; the patient is fully unconscious.

Did you see what I did differently this time? Yeah, just trying to keep myself awake.

c)

19. 3, 6, 12, 24, … Oooh! Geometric, common ratio of 2.

a) 48, 96, 192, 384. I was tempted to get a calculator out for that last value, but decided against it.

b) = 3 · 2n – 1 Note that there is nothing I can do to

simplify that. They are not powers of the same base, and the power is only on the 2, not the 3.

c)

20. 4, 12, 36, 108, … Geometric, r = 3.

a) 324, 972, 2916, 8748. Yes, I did those in my head, but before you see this file, I will have checked my answers in the back, and you’ll have no idea whether I initially got them right.

b) = 4 · 3n – 1

c)

21. 5, –5, 5, –5, … Finally! One I like! At first it just looks

weird, but when you consider that the instructions said that all the sequences here are either arithmetic or geometric, this must be geometric with a common ratio of –1. a) 5, –5, 5, –5. Aw, yeah. b) Did you think that there was

something magic about a and u? There’s not.

c)

( )1 1nu u n d= + -

1

1

811, 2n n

uu u n-

= -ìí = + ³î

( )1 1nu u n d= + -

1

1

10.050.1, 2n n

uu u n-

=ìí = - ³î

( )1 1nu u n d= + -

1

1

1001, 1n n

uu u n-

=ìí = - >î

1 52, , 1, ,2 2

- - !

32

-

8 11 14 17, , ,2 2 2 2

- - - -

( )1 1na a n d= + - ( ) 32 12

n æ ö+ - -ç ÷è ø

3 3 7 322 2 2 2n n= - + = -

1

1

23 , 12n n

a

a a n-

=ìïí

= - >ïî

11n

nu u r -=

1

1

32 , 1n n

uu u n-

=ìí = >î

11n

na a r -=

1

1

43 , 1n n

aa a n-

=ìí = >î

( ) 111 5 1 nn

nb b r --= = × -

1

1

5, 1n n

bb b n-

=ìí = - >î

22. 3, –6, 12, –24, … Geometric, multiplying by –2.

a) 48, –96, 192, –384. b)

c)

23. 972, –324, 108, –36, … This is definitely easier to figure out from

the last two pairs than the first one. It’s dividing by –3, which means that the value

of the common ratio is .

a) 12, –4, . It gets small pretty

fast.

b)

c)

24. Five arithmetic means between 15 and –21

is asking for the five missing terms in the arithmetic sequence that goes like this:

15, ___, ___, ___, ___, ___, –21. If you count there, you’ll see that 15 is the

first term (duh) and the seventh term is –21. We can use the formula for the nth term of an arithmetic sequence to find d, the common difference, and once we have that, filling in the blanks will be easy.

Then the missing terms are found by adding –6 repeatedly: 9, 3, –3, –9, –15. You can check to make sure you’re right by adding it once more. If you get –21, you’re a winner.4

4 I’m going back to check my answers with BoB, and he has also given 15 and –21 on either end of the list. Those

25. Three arithmetic means between 99 and 100 means that term number 1 is 99 and term number 5 is 100.

Then the missing terms are

.

26. Ah, this time we have rather than . If it were the first term, then we could do exactly the same process as the previous problem to get the value of the common difference, and thus find the formula for the nth term. In this case, there are two possible ways to proceed. The oh-so-clever way is to make those terms the first and tenth of a new sequence, say , and then do it like the problem before. I’ll do it the less clever, more straightforward way.

If , then becomes . If , then

becomes . Those two equations together form a system

I can solve: . While I might

choose to use the linear system solver on my calculator, this particular system solves very nicely by elimination, since the terms cancel when the equations are subtracted. Subtracting the top equation from the bottom (to keep the numbers positive) gives 36 = 9d, and d = 4. And then

means . So the

formula we want is = 3 + 4n – 4 = 4n – 1.

aren’t actually the arithmetic means, and I’m not including them in my answer.

( ) 111 3 2 nn

nu u r --= = -

1

1

32 , 1n n

uu u n-

=ìí = - >î

13

-

4 4,3 9-

11

119723

nn

nu u r-

- æ ö= = -ç ÷è ø

1

1

9721 , 13n n

u

u u n-

=ìïí

= - >ïî

( )( )

1

7 1

1

7 121 15 636 6

6

nu u n d

u u dd

dd

= + -

= + -

- = +- == -

( )( )

1

5 1

1

5 1100 99 41 4

14

nu u n d

u u dd

d

d

= + -

= + -

= +=

=

1 1 399 , 99 , 994 2 4

3a 1a

nb

3 11a = ( )1 1na a n d= + -

111 2a d= + 12 47a =

( )1 1na a n d= + - 147 11a d= +

1

1

11 247 11

a da d

= +ìí = +î

1a

( )111 2 4a= + 1 3a =

( )3 1 4na n= + - ×

27. And this is just like the previous one. I think this time I’ll take the other method. Define a new sequence, with terms called b. Each term in b has a term number six less than the corresponding term in a. That means since , and since

Got it?

–10 = –48 + 6d 38 = 6d

d = .

Unfortunately, I still have to find , since I need a formula for , not . But if

, then

–48 = + 38, and = –86.

So .

28. Geometric means this time! Those are

missing terms in the geometric sequence; we’re looking to fill in the blanks in 7, ___, ___, ___, ___, 1701. It just means we need a different formula: . Literally counting on my fingers5 here, I discovered that 1701 is the sixth term.

1701 = 7r5 243 = r5 Then to get the geometric means, we just

multiply by 3 repeatedly: 21, 63, 189, 567. Just as a check I multiplied the last one by 3, and I did get 1701, so that must be correct.

29. Now we’re asked for just one geometric

mean between 9 and 64. You can do exactly

5 Yes, counting on my fingers. And I’m not ashamed. I got the answer right, didn’t I?

the same thing as before; 9 is the first term and 64 is the third.6

Important note: Back in geometry class, you used the geometric mean as the length of a segment, and it was always positive. Here, there is no such restriction.

Then the missing term is either

, or .

30. Given the first term (24) and the third (6) of

a geometric sequence, find the fourth term and an expression for the nth term. Really, this requires exactly the same sort of work as the previous problem up until we find the value of r. Afterwards, we use it differently.

Since there are two values here, I’ll expect to find two different answers for each of the questions asked. They should only differ by that sign, though.

The fourth term can be found from the

third: or . The

nth term comes from the formula.

or .

If you’re wondering if geometric questions will always have two sets of

6 It’s also possible to do this the way you learned back in geometry class, with a proportion. One geometric mean

would be the solution to the proportion .

7 148, 48a b= - = -

13 710, 10.a b= - = -

( )1 1nb b n d= + -

( )7 1 7 1b b d= + -

38 19 166 3 3= =

1a

na nb

( )7 1197 13

a a= + - ×

1a 1a

( ) 19 19 186 1 923 3 3na n n= - + - × = -

11n

nu u r -=

6 16 1u u r -=

5 243 3r = =

3 13 1

2

2

64 964983

u u rr

r

r

-=

=

=

= ±

89 243× =

89 243

× - = -

3 13 1

2

2

6 241412

u u rr

r

r

-=

=

=

= ±

416 32

u = × = 416 32

u = × - = -

11

11242

nn

nu u r-

- æ ö= = ç ÷è ø

11242

n

nu-

æ ö= -ç ÷è ø

9

64

x

x=

answers like this one did, the answer is no. If there had been an even number of missing terms, the power on r would have been odd. In that case there’s no plus-or-minus involved, and just one solution.

31. Common ratio , fourth term . Find the

third term. I totally started to do a whole bunch of ugly fraction work here with the nth term formula before I came to my senses. If I have the fourth term, and I want the third, perhaps I should just divide by the common ratio.

. Done.

32. “Which term” is another way of asking you

to find the value of n in the nth term. We know the first term, the nth term, and it’s clear from the first two terms that r = 3.

At this point, you could use logarithms, provided you remember enough about them for that, but we actually know that the answer will be an integer; a little intelligent guess-and-check will tell you that the power on 3 is 9. So n – 1 = 9 means n = 10.

33. Now this is more interesting. If a particular

value is a term of a geometric sequence, there will be an integral7 value of n that makes the nth term formula give that value. What I’m saying, not especially well, is that we need to repeat the process from the previous problem, but we may find that n does not come out pretty; if so, the fraction there would not be a term in the sequence.

and

7 That’s the adjective form of integer. It also turns out to have another really important mathematical meaning, but that’s for calculus class.

Here’s another way to think about finding r: to get from , you have to multiply by r three times (count on your fingers if you want, I’ll wait). That means .

It will also be helpful to figure out the first

term; since , .

Now, if is a term of the sequence,

the equation will have a

solution that gives an integral value of n.

Again, I could use logarithms, but since we haven’t done that yet this year, I’m not going to. I’m just going to recognize that 512 is a power of 2. In fact, it’s 29, and that fact is in my brain. You can double repeatedly or use a calculator to figure it out. So if the denominator has to be raised to the ninth power, the numerator does, too. Here’s the moment of truth: my calculator says that 39 = 19683. And therefore n – 1 = 9, n = 10, and the answers to the questions are, “Yes, it is” and “the tenth one.”

34. The investment of €2500 is for 10 years at

4% compounded semi-annually. The formula for compound interest is

. Here, P = 2500, t = 10,

r = 0.04, and n = 2, for twice a year.

37

143

14 3 14 7 983 7 3 3 9÷ = × =

11

1

1

137781 7 319683 3

nn

n

n

u u r -

-

-

=

= ×

=

3 18u = 62434

u =

3 6 to u u

36 3u u r=

3

3

243 18427832

r

r

r

=

=

=

3 13 1u u r -= 3

1 22

18 832

uur

= = =æ öç ÷è ø

1968364

11

1968364

nu r -=

119683 3864 2

n-æ ö= ç ÷è ø

119683 3512 2

n-æ ö= ç ÷è ø

1ntrA P

næ ö= +ç ÷è ø

≈ €3714.87, to the

nearest euro cent.8 On an exam, the instructions would likely say exact or three significant figures unless otherwise specified for numerical answers. With money, however, they do often specify something like the nearest whole unit. Since this set of problems said nothing in particular about precision, I’m going with what I know to be the standard number of decimal places for the currency. It looks like BoB is doing that, too.

35. Same formula again, but this time P = 1000,

r = 0.06, n = 4, and t = 18.

≈ £2921.16.

36. This one goes the other way; A is given as

€3000, and we have to work backward to find P. Again, r = 0.06, n = 4, and now t = 6.

, so

≈ €2098.63.

37. To find the sum of the arithmetic series 13 + 19 + … + 367, there are two formulas

to choose from. One uses the first and the nth terms, and the other uses the common difference instead of the nth term. Both of them require the value of n — that is to say, the number of terms being added. So it would be wise to start with that. (The common difference, by the way, is 6. You can see that, can’t you?)

In order to determine the number of terms, the nth term formula is a good

8 Yes, a euro cent is the standard subdivision of that currency, but in English (and most of the languages of countries that use it), it’s officially just called a cent. I’ll call it that from now on.

choice. We know that the first term is 13 and the nth is 367.

367 = 13 + (n – 1) · 6 = 13 + 6n – 6 367 = 6n + 7 360 = 6n n = 60. And now I can choose whichever of the two

sum formulas I’d like. I pick the one with the nth term, since I know it already.

38.

Well, just a glance suggests that the numerators are doubling and the denominators are tripling, suggesting a geometric series. If you look more carefully, you’ll see that the signs of the successive terms alternate. That means the common

ratio must be . But there’s another

catch. To find the sum of a finite geometric series, we need to know the number of terms involved. The first numerator is 21, the second is 22, and so on. So what power of 2 is 4096? Using my math brain, I know it’s 12. Go ahead, check it if you don’t believe me.9 Therefore the number of terms, n, is 12.

I used a calculator for that, as any sane person would.

9 I’m not one of those people who can do big numbers in her head. I just remember that 210 starts with 10. It’s 1024. And doubling twice more makes 4096. So that must be 212.

2 100.042500 12

A×

æ ö= +ç ÷è ø

4 180.061000 14

A×

æ ö= +ç ÷è ø

4 60.063000 14

A×

æ ö= +ç ÷è ø

2430000.0614

A =æ ö+ç ÷è ø

( )1 1nu u n d= + -

( )

( ) ( )

1

60

260 13 367 30 380 114002

n nnS u u

S

= +

= + = =

4 8 16 40962 ?3 9 27 177147

- + - + - =!

23

-

( )1

12

12

11

22 13 210,938 1.192 177,14713

n

n

u rS

r

S

-=

-æ öæ ö- -ç ÷ç ÷ç ÷è øè ø= = »

æ ö- -ç ÷è ø

39. To evaluate , it would be

helpful to know what kind of series it is. While one can look at the kth formula there (also known as the summand) to see that the underlying sequence is arithmetic, I generally find that it’s very helpful to write out the first few terms. I’ll do that here.

If k = 0, 3+ 0.2(0) = 3. If k = 1, 3 + 0.2(1) = 3.2. If k = 2, 3 + 0.2(2) = 3.4. If I were to go all the way to k = 11, I’d

have 3 + 0.2(11) = 5.2. So my series is 3 + 3.2 + 3.4 + … + 5.2. The

terms increase by 0.2, so it’s arithmetic, and there are 12 terms. (Wait, not 11? Nope, because you have to count the term numbered 0. That’s 12 total.)

Then , so

.

40. looks ridiculously

similar to problem 38. In fact, I just copied it from there and backspaced out the last bit. The difference is that this one never stops; it’s an infinite geometric series. That actually makes the problem easier than before, which is a little counterintuitive — how can adding up infinitely more terms be simpler? Mathemagic!

, provided that < 1. In this case,

, so the formula should work just

fine.

, or 1.2

exactly.

41. Okay, this one is a little ugly.

It’s infinite, which means if I hope to get a value in this course, it had better be geometric.10 So how do I find the common ratio? By remembering what a common ratio is! It’s what you get when dividing each term by the one before it. Therefore

. I could write that

as a fraction under the radical if I chose, but I’m okay with this version. If you look at the sequence again, you can see that this value clearly works to get from the third term to the fourth. It’s a little more difficult to see how it gets you from the second to the third term, or the fourth to the fifth. Go ahead, see if you understand. I’ll wait….11

Okay, then, here’s the sum formula:

Oh, dear. I shall attempt to simplify. It’s one of those complex fractions, and multiplying by the common denominator of the little fractions over itself (to make 1) should help some, but I’m not sure the answer will be very nice.

10 There are other sorts of infinite series that have finite sums one might determine, but they generally require calculus, I think. I cannot think of an example that doesn’t. 11 No, four dots isn’t a typo. It’s an ellipsis followed by the period to end the sentence. So there.

( )11

03 0.2

kk

=

+å

( )12n nnS u u= +

( ) ( )1212 3 5.2 6 8.2 49.22

S = + = =

4 8 1623 9 27

- + - +!

1

1uSr

=-

r

23

r = -

2 2 3 6252 5 5133

S = = = × =æ ö- -ç ÷è ø

1 2 1 2 22 3 92 3 3 3+ + + + +!

22 2 22 3

1 12 3 32

r = = × =

1

12

1 213

uSr

= =-

-

1 2 32 32 2

2 2 3 2 2 31 2 33 33

2 3 2 2

× =×

- -

=-

If you were a real obsessive, you could rationalize the denominator. I’m not.12

42. Changing repeating decimals to fractions

can be really easy in some circumstances, and you’re welcome to use any special tricks you know to help. This unit is about sequences and series, though, so I’m going to use infinite geometric series to do them. a)

And now you can probably see the infinite geometric series. The first

term is , and the common ratio is

, which will add a zero on the

bottom with each successive multiplication (or, equivalently, move the decimal over one place).

The sum: .

b) = 0.3 + 0.045 + 0.00045 + 0.0000045 + …

As you can see, to move the decimal over two places at a time, I need to increase the number of zeros in the denominators by two at a time.

This time, the geometric series doesn’t

start with , though. It’s just the

part. So

I’ll calculate that, and then add the

on at the end of the process. 12 Okay, some of you are:

. No, in no way

do you have to do that on the test.

For the infinite geometric series part,

And finally,

.

Fun, right? c) What fun! This time the repeating

starts after one additional term. I think I’m going to use decimals for a change. I suspect by now you understand what’s going on with the fractions.13

So the geometric part starts with

0.0029 for a first term, and has a common ratio of 0.01 to move the decimal point two places at a time.

I suppose if you wanted an improper fraction rather than a mixed number,

you could change it to . I do not

feel inspired to put a question like this on the test, though.

43. The binomial theorem: Find the coefficient

of x6 in the expansion of (2x – 3)9. I’ll do this one by setting up the expansion of the entire binomial and then picking the term I need. Next time I’ll use the formula to go straight to the term I need. Both processes are legitimate, and you can choose whichever one seems best for the problem you’re working on at the time.

13 I have just noticed that I’m on page 9, but only question 42 of 63. I hope some of you have made it here with me.

0.7 0.7777...=0.7 0.07 0.007 0.00077 7 7 710 100 1000 10000

= + + + +

= + + + +

!

"

710

110

1

7 7710 10

1 91 9110 10

uSr

= = = =- -

0.345 0.3 0.045= +

3 45 45 4510 1000 100000 10000000

= + + + !

310

45 45 451000 100000 10000000

+ + !

310

3 2 3 2 2 2 9 2 6 6 2 6 6 2 6 3 612 8 4 22 3 2 2 2 3 2 2 4 9 4 4

+ + + + +× = = = =

-- + -

1

45 451000 10001 991 1100 100

45 100 5 11000 99 110 22

uSr

= = =- -

= × = =

3 1 33 5 380.34510 22 110 110 110

= + = + =

3.2129 3.21 0.0029 0.000029 ...= + + +

0.0029 0.00293.21 3.211 0.01 0.9921 0.0029 21 293 3100 0.9900 100 99002079 29 2108 5273 3 39900 9900 9900 2475

+ = +-

= + + = + +

= + + = =

79522475

To expand the ninth power, I’ll need the numbers from the row of Pascal’s triangle that starts 1, 9, …. That row14 looks like

1, 9, 36, 84, 126, 126, 84, 36, 9, 1. The powers on the first term, 2x, start at 9 and work themselves down to 0, while the powers on –3 start at 0 and work up to 9, like this:

1(2x)9(–3)0 + 9(2x)8(–3)1 + 36(2x)7(–3)2 + 84(2x)6(–3)3 + … Wait! I’m looking for the term with x6, right? I am NOT typing any more of that! I have the term I need right here. It’s 84(2x)6(–3)3 = 84 · 64x6 · –27 = –145,152x6. The coefficient is –145,152.

44. Okay, for the coefficient of x3b4 in (ax + b)7,

I’ll take the other tack. The general term of the expansion of the binomial (a + b)n is

. In this case, instead of a, we

have ax, and n = 7, so this becomes

. I’m looking for the term that

contains x3b4. If you look carefully, you can conclude that r = 4. So I just need to work

out = 35a3x3b4, and the

coefficient I need is 35a3. (Note: I got the 35 on the calculator by typing ncr(7,4). You’re responsible for knowing how to do that.)

45. Find the constant term of .

What’s a constant term? Using the crazy logical vocabulary that actually works a lot of the time in math, it’s the one that’s not changing — the one with no variable in it. Another way to think about it is that it’s the term where z is raised to the zero power.

I’ll go with the same method I used in problem 44: the general term of the expansion of the binomial (a + b)n is

14 I got it by clever use of a function and tables on my TI-nspire. You could do the same with an ncr(9, x) command.

. In this case, a = , b = –z, and

n = 15. So the general term becomes

. To make this into

something more useful, I’ll do some algebra on the powers.

Whew. Did you catch what I did? The 15 – r became the power on both numerator and denominator of that fraction, and then since I was dividing two powers of the same base, I subtracted the exponents. But how does this help? Don’t worry, I’ll remind you. We are looking for the constant term, the one with a power of zero on z. That will mean that we can set 3r – 30 = 0 and know that r = 10. Finally, we just put in that value of r and work out the answer.

46. Expand (3n – 2m)5. This time I want all the

terms. I could generate them individually with that formula from the last problem, but I won’t. I’m just going with my old buddy Pascal.

The row of Pascal’s triangle I need here is one I’ve memorized, because it’s easy: 1, 5, 10, 10, 5, 1. The powers on 3n will start with 5 and decrease as I go across left to right, while the powers on –2m will start at 0 and increase. Be careful where you put the parentheses and the negative sign. Misplacement of those is the most common error I see on problems of this sort.

n r rna b

r-æ ö

ç ÷è ø

( )77 r rax br

-æ öç ÷è ø

( )7 4 474ax b-æ ö

ç ÷è ø

15

2

2 zz

æ ö-ç ÷è ø

n r rna b

r-æ ö

ç ÷è ø

2

2z

( )15

2

15 2 rrz

r z

-æ öæ ö -ç ÷ç ÷è øè ø

( ) ( ) ( )

( )

( ) ( )

( )

15 15

2 2 15

15

30 2

30 215

15 3 30

15 152 2 1

15 2 1

152 1

152 1

r rr r r

r

rr r

r

r r rr

rr r

z zr rz z

zr z

zr

zr

- -

-

-

-

- --

- -

æ ö æ öæ ö - = -ç ÷ ç ÷ç ÷è øè ø è ø

æ ö= -ç ÷è øæ ö

= -ç ÷è øæ ö

= -ç ÷è ø

( )1015 10 310 30

0

152 1

10

3003 32 1 96,096

z

z

- × -æ ö-ç ÷

è ø= × × × =

1(3n)5(–2m)0 + 5(3n)4(–2m)1 + 10(3n)3(–2m)2 + 10(3n)2(–2m)3 + 5(3n)1(–2m)4 + 1(3n)0(–2m)5 = 1 · 243n5 · 1 + 5 · 81n4 · –2m + 10 · 27n3 · 4m2 + 10 · 9n2 · –8m3 + 5 · 3n · 16m4 + 1 · 1 · –32m5

= 243n5 – 810n4m + 1080n3m2 – 720n2m3 + 240nm4 – 32m5.

If you think that’s hard to read, try typing it. 47. Find the coefficient of r10 in (4 + 3r2)9.

Okay, this time I’m skipping straight to the term I want. I’m tempted to go with the “counting on my fingers” method, but I know that is likely to confuse some of you, especially as I do not intend to make a video showing how I do the counting. So I’ll use the general term formula again.

The general term of the expansion of the

binomial (a + b)n is . In the given

binomial, a = 4, b = 3r2, and n = 9. Since I want the term with r10, the exponent on 3r2 must be 5. (I know, it’s annoying that there’s an r in the formula and an r in the binomial. Suck it up, you’re in precalculus.)

, and the

coefficient is 126 · 256 · 243 = 7,838,208. 48. If this problem sounds like ones we’ve done

earlier, it is, just with more words. The sequence is arithmetic, and we know the first and fourth terms. That should be enough to find the common difference.

Yes, I did that last step in my head. You can, too, right? Now that I have d, knowing that the nth term’s value will allow me to find n.

49. Totally unrealistic question about studying

here. In fact, it’s very similar to an old exam question, but with a different setting. Nick studies 12 hours in the first week and increases his time by 2 hours per week; Charlotte also starts by studying 12 hours per week, but increases her time by 10% each week. a) In week five, Nick’s arithmetic pattern

gives , or

hours. As she increases her time by 10% a week, her common ratio is 1.10, and Charlotte’s geometric pattern gives , or , or about 17.6 hours (to 3 s.f.).

b) The total number of weeks is the sum of a series.

Nick:

= 390 hours

Charlotte:

≈ 381 hours (3 s.f.)

c) For Charlotte to exceed 40 hours per week, we need her nth term to be greater than 40.

. This could be solved by logarithms, but the introduction of the inequality changes things a bit — for instance, does the inequality sign change directions when you take a logarithm of both sides? How do you know? A simpler idea is to use technology, either by making a table, or by graphing. I’m

n r rna b

r-æ ö

ç ÷è ø

( )59 5 2 1094 3 126 256 243

5r r-æ ö

× = × ×ç ÷è ø

( )( )

1

4 1

1

4 119 4 3

5

nu u n d

u u dd

d

= + -

= + -

= +=

( )( )

1 1

99 4 1 599 4 5 5 5 15 100

20

nu u n d

nn n

nn

= + -

= + - ×

= + - = -==

( )1 1nN N n d= + - ×

( )5 12 5 1 2 20N = + - × =

11n

nC C r -=5 1

5 12 1.10 17.5692C -= × =

( )( )12 12nnS N n d= + -

( )15 2 12 14 22

= × + ×

( )1 11

n

n

C rS

r-

=-

( )1512 1 1.11 1.1-

=-

1 11 12 1.1 40n n

nC C r - -= = × >

going with a table. I will first define a function representing the nth term, then have the calculator generate a list of values.

The define command is in the Actions

menu. Then add a new lists and spreadsheets page.

Switch that page to table view.

Choose the name of the relevant

function. (Hint: I just defined it.)

And then scroll down to find the

relevant value.

So she first exceeds 40 hours of study

in week 14. d) Well, the equation I’d like to solve is

Nn = Cn, or . This, however has no analytical solution. Logarithms will not do it for you. There are only approximate methods of solution, and since I already have a table for Charlotte, I’m just going to add one for Nick.

I wasn’t allowed to use the same letter

for the function and the variable, and the TI-nspire doesn’t distinguish between capital and lower-case (a “feature” which continues to annoy me).

Then a click at the top of the second column of the table lets me choose the function to display there.

And then we look for when Charlotte

catches up to Nick.

Week 12.

A lesson here: being able to solve problems analytically is absolutely important. However, it also is essential to know when that’s not possible (or maybe just not feasible) and then how to go about solving them with the help of technology.

50. These diet plans sound all science-y, what

with their use of “grams,” but they also seem like those ads in the magazines that your grandmother has at her house. I wouldn’t trust them. Plan A has an initial loss of 1000 g, with an increase of 80 g per month, so it’s an arithmetic sequence. Plan B starts out the same, but increases geometrically with a common ratio of 6%. a) Remember that “write down” doesn’t

require showing any work. However, I’m going to show it anyway, in case anyone needs to see it.

B1 = 1000 g B2 = 1000 · 1.06 = 1060 g B3 = 1000 · 1.062 = 1123.6 g b) The weight lost in the twelfth month is

the same as the twelfth term of each sequence.

g

or 1900 g (to 3 s.f.)

c) The total weight loss would be the sum of the appropriate series.

i) Plan A: =

17280 g, or about 17.3 kg

( ) 112 1 2 12 1.1nn -+ - × = ×

( )12 1000 12 1 80 1880A = + - × =12 1

12 1000 1.06 1898.3,B -= × »

( )1212 1000 18802

S = +

ii) Plan B: ≈

16869.9, or 16.9 kg (3 s.f.) 51. If I’m reading this correctly, you’re

investing €500 at the beginning of each of the next ten years, and it’s earning interest at 6% compounded annually. That’s what’s called an annuity, and there’s a formula. You do not have to memorize it. I looked it up, too, in §3.4. a) For how much the first €500 is worth,

I don’t need an annuity formula, though. It’s just a geometric sequence, with r = 1.06, the multiplier that will add 6% to the value every year. The tricky bit is actually related to the fact that you’re making the investment at the beginning of each year, and you want the value at the end of year 10. That means the power on 1.06 is 10, not 9.

≈ €895, to the nearest euro.15

b) , where FV

represents the future value of the annuity (which we would like to find), R is the periodic payment (€500), i is the interest rate per period (0.06 in this case), and m is the number of periodic payments (10).

≈ €6986, to

the nearest euro. Depending on when you read this, that’s probably something under $10,000. And we’re talking about you buying the car 10 years from now. You might want to step up the savings a bit; you’re going to want a nicer ride than that.

52. An arithmetic sequence begins 6, 9.5, 13.

The common difference is 9.5 – 6 = 3.5. 15 I have discovered that BoB gave his answers to the nearest cent, proving that he doesn’t always read directions.

a) Find the 40th term. = 142.5 b) For the sum of the first 103 terms,

there are two formulas to choose from. Since I do not already know the value of the 103rd term, I’ll choose this one:

=19003.5

53. The runner starts with a 2 km run, and

increases that by 500 m (that’s 0.5 km) each day she trains. It’s an arithmetic sequence. a) She will first run a distance of 20 km

when the nth term equals 20. 20 = 2 + (n – 1) · 0.5 = 0.5n + 1.5 18.5 = 0.5n n = 37. So that’s on day 37. b) By day 37, the total distance she will

have run is the sum of an arithmetic series with 37th term 20.

= 407 km. 54. The nation of Telefonica is located in the

Indian Ocean, near the Maldives.16 a) The common ratio of a geometric

sequence is the ratio of successive

terms. .

b) If this ratio holds, 2012 will be the year when n = 13. Yes, 13. If the year 2000 is n = 1… count it on your fingers. Got it?

≈ 207594, to the nearest whole Telefonican.

c) The number of new participants would first exceed 50,000 when un is at least that large. I feel a table coming on.

16 That’s a lie. It’s really near the Galapagos, off the coast of South America.

( )12

12

1000 1 1.061 1.06

S-

=-

1010 500 1.06u = ×

( ) 11 11

miFV R

i

+æ ö+ -= -ç ÷

ç ÷è ø

111.06 1500 10.06

FVæ ö-

= -ç ÷è ø

( ) ( )1 1 6 40 1 3.5nu u n d= + - = + - ×

( )( )12 12nnS u n d= + -

( )( )103 2 6 103 1 3.52

= × + - ×

( )1 1nu u n d= + -

( ) ( )137 2 20

2 2n nnS u u= + = +

2400 3600 1.51600 2400

r= = =

13 113 1600 1.5u -= ×

Then the table.

Looks like year 10, which is 2009. Go

ahead, count on your fingers again… d) The total number of participants

between 2000 and 2012 would be the sum of the first 13 terms of the series.

≈ 619582

Telefonicans. I’ve gone back and checked answers in the back, and BoB has apparently rounded up both here and in part (b). I disagree with his decision. These formulas are just models, and regular, ordinary rounding rules apply. Less than 0.5 should round down.

e) If the total adult population stays near 800,000, then in a very short while (one more year) every adult will have a cellular plan, and within a couple of years after that, all of their children and pets will, too. At some point, the growth is just not sustainable.17

55. I feel like I just did this problem, when it

was called number 48. Wait! This is clearly a problem about martial arts — it has a “fist term.” With the “fist” and fourth terms, we can find d.

–12 = 3d, and d = –4 Once we know d, the nth term will allow us

to find the value of n using the same formula.

17 Maybe they can import people from the favelas of Rio de Janeiro. They’d just have to learn to speak Telefonican, which is, after all, related to Portuguese. It shouldn’t be that hard.

56. Pretty picture! The midpoints of the sides of

a square are joined. And then the midpoints of the sides of that square, and then of that one, and then… a) I wish that they had named the

vertices of the outer square. I’m going to call the square DAVE. His vertices will be named as you can see in the picture here.

Now, finding the length of a side of

MNPQ is equivalent to finding the hypotenuse of ∆PEN, for instance. Since the sides of DAVE are each 1

cm long, . Because

ÐPEN is an angle of a square, it measures 90°, and ∆PEN is an isosceles right triangle. The ratio of its sides is 1: 1: , and I conclude that the length of the hypotenuse, NP, is

. Q.E.D.18

b) The area of a square is the square of the length of one of its sides (and that’s why it’s called “squaring” the number).

.

c) i) The length of a side of RSTU is the length of the isosceles right

18 That’s the abbreviation of “quod erat demonstrandum,” which is Latin for, “Aw, yeah, told you I could do it.”

( )13

13

1600 1 1.51 1.5

S-

=-

( )( )

4 1 1

13 25 4 1 3 25

u u n d

d d

= + -

= + - = +

( )11995 25 1 4 4 29n n- = + - ×- = - +

11995 29 30064

n - -= =

-

12

PE NE= =

2

1 222 2× =

22 2 12 4 2

æ ö= =ç ÷ç ÷

è ø

D A

V E

triangle ∆PUT. A leg of that triangle, PT, is half of PN (remember, we’re joining midpoints). So

. The

length of the hypotenuse of ∆PUT, TU, is times as long as the leg:

.

And then the area of square

RSTU is .

ii) In every case, we will continue to see the pattern that the side of

the next square is of the

previous square, and thus the area of the next square is

of the

previous square. The sequence of the areas is therefore geometric with a common ratio

of .19

d) i) The area of the tenth square will be the tenth term of the sequence with first term 1 (the area of the outer square, DAVE) and

common ratio .

ii) The sum of the areas of all the squares is the sum of the corresponding infinite series:

.

19 You can do this with pictures, too. It’s easier to see if you don’t show all the inner squares, just one of them. Wow. Nineteen footnotes.

57. Tim is obsessive-compulsive, and if he does

not swim precisely 20 meters more each week than the previous week, he may have a breakdown. Tim also writes math books for a living, and bases some of the problems in them on his own life. This problem is an arithmetic sequence. a)

= 1220 m in the fifty-second week.

b) His total mileage20 is the sum of those

52 terms:

= 36,920 m, or

36.92 km. (Or about 22.9 miles, if you’re looking for mileage.)

58. So in each case, of a square is shaded.

Then each of the eight unshaded parts is divided into nine equal pieces and one of those is shaded. And so on. a) Since we know the original area of the

square is 1, the area of A is of that,

or . Then the area of B is of a

square congruent to A, which makes

its area .

b) The small squares in the third

diagram are the size of B, which

makes them .

c) The total shaded area in the second

diagram is ,

and the total shaded area in the third

20 “Meterage”? Can you say mileage when you’re measuring in the metric system?

1 1 2 22 2 2 4

PT NP= = × =

2

2 4 2 124 4 4 2

TU = × = = =

( )2

2 1 12 4

TU æ ö= =ç ÷è ø

1 22×

21 1 12 22 4 2æ ö× = × =ç ÷è ø

12

12

10 1

10 9

1 1 112 2 512

u-

æ ö= × = =ç ÷è ø

1 1 21 112 2

S = = =-

( )1 1nu u n d= + -

( )52 200 52 1 20u = + -

( )12n nnS u u= +

( )5252 200 12202

S = +

19

19

19

19

1 1 19 9 81× =

19

3

1 1 1 1 19 9 9 9 729æ ö× × = =ç ÷è ø

1 1 9 8 1789 81 81 81 81+ × = + =

diagram is =

.

d) In order to figure out what happens in the end, I’ll need to get a better numerical description of the pattern. It’s looking geometric because of all

the multiplication. The first term is .

For the common ratio, I should divide

successive terms. . This

also makes sense geometrically. We took the eight copies of original shape, and shaded one-ninth of each of them.

The infinite series therefore has first

term and common ratio , for a

sum of . That’s the

shaded part. And therefore the unshaded part is… zero? Yep, in the limit, the whole thing would end up shaded. You can kind of see how it’s starting to turn solid blue by the third one.

59. I don’t expect you would copy the table; on

a test, you’d just write in the blanks. So I’m not going to copy the table, either. a) i) This one is neither arithmetic nor

geometric. There’s certainly a pattern, but it’s “times ten, then plus two,” and that is some crazy mutant hybrid.

ii) The pattern here is “times ,”

which makes the series geometric. As |r| < 1, it converges, too.

iii) In each case, the terms decrease by 0.02, making this one arithmetic.

iv) And in this one, the multiplier is

, so it’s a divergent geometric

series. b) The sum of an infinite geometric

series with first term 2 and common

ratio is .

60. Two companies have different salary schedules. Kell’s is arithmetic, with starting value of €18,000 and common difference of €400; YBO’s is geometric with a starting value of €17,000 and common ratio of 7%. a) i) Kell: second year is €18,400;

third year is €18,800. YBO: 17000 · 1.07 = €18,190 in

the second year; third year is 17000 · 1.072 = €19.463.30. As precision is not specified, I went to the nearest cent.

ii) The total amount earned in 10 years would be the sum of a series.

Kell:

= €198,000. YBO:

≈

€234,879.62 iii) And this would be the tenth term

instead. Kell:

= €21,600 YBO: ≈

€31,253.81. Sorry you took that job at Kell, huh?

b) i) Tim, the swimmer, works at Kell. Merijayne works at YBO. Merijayne would start earning more than Tim when

gets bigger than

1 1 18 8 89 81 729

æ ö+ × + × ×ç ÷è ø

81 72 64 217729 729 729 729

+ + =

19

88 9 881

1 81 1 99

= × =

19

89

1 19 9 18 119 9

S = = =-

23

43

23

1 2 2 62 11 13 3

uSr

= = = =- -

( )( )1010 2 18000 10 1 4002

S = × + - ×

( )10

10

17000 1 1.071 1.07

S-

=-

( )10 18000 10 1 400u = + - ×

10 110 17000 1.07u -= ×

117000 1.07n-×

18000 + (n – 1) · 400. And logarithms would be no help, so stop assuming you need them!21 A graph or a table would work. Since I am convinced you aren’t comfortable enough with tables yet, that’s what I’m going to do.

MJ is earning more by year 3. ii) And for their total earnings,

we’d just compare the sum formulas. For Tim, it’s

and for MJ, it’s

.

Her total is higher by year 4.

61. The theater has 24 rows of seats: 16 in the first row, and two more in each successive row. a) In row 24, this would be the 24th term

of an arithmetic sequence with first term 16 and common difference 2.

seats

21 Okay, I know that not all of you were thinking that. But I bet some of you were. As wonderful as logs are, they won’t fix everything involving exponents.

b) The number of seats in the entire theater is the sum of the 24 terms in the arithmetic series.

seats

62. €7000 is invested at 5.25% interest, initially

compounded annually. a) There’s a formula for compound

interest in general. You could also just use the nth term formula for a geometric sequence, but you probably need more practice with the interest formula.

b) For this to become €10000, we’d need to solve 7000 · 1.0525t = 10000. And logarithms would totally work. Knock yourself out. Here’s a graph.

It looks like the answer is 6.97, but

that’s totally not true. The interest is only being compounded at the end of the year. So the answer is 7 years.

c) Ah, would you be better off if your interest were only 5%, but it were compounded four times as often. The formula becomes

, and I’ll just add

that to the earlier graph to see how they compare. With the same window settings, it was a little hard to see, so I zoomed in near the place where both curves get to €10,000.

( )( )2 18000 1 4002nnS n= × + - ×

( )17000 1 1.071 1.07

n

nS-

=-

( )24 16 24 1 2 62u = + - × =

( )2424 16 62 9362

S = + =

1ntrA P

næ ö= +ç ÷è ø

10.5257000 1 7000 1.05251

ttæ ö= + = ×ç ÷

è ø

40.057000 14

t

A æ ö= +ç ÷è ø

The blue curve is the original (I just

realized that the graph is set to show 3 significant figures, but I promise that the graph truly uses 1.0525, not 1.05, for the multiplier). The black one is the new one, and it arrives at €10000 two-tenths of a year later. So, no, it wouldn’t be an improvement (not just for the same number of years as the first investment, but ever).22

63. Ah, I like this one! We have information

about the sums of terms, rather than the terms themselves. It would be way more interesting if it weren’t the first two partial sums that were given, but at page 18 of the answers, I’ll take what I can get. a) If , then , too. It’s the

sum of the first one term (which is pretty silly to type or read). Then since , we can conclude that the second term must be 11.

b) If the first term is 9 and the second is 11, the common difference is 2.

c) And the sequence goes like this: 9, 11, 13, 15. The fourth term is 15.

22 BoB disagrees with this answer. I’m convinced he’s wrong; (1+0.05/4)4 ≈ 1.05095, which is less than 1.0525.

1 9S = 1 9u =

2 1 220S u u= = +