math 140 quiz 6 - summer 2004 solution review (small white numbers next to problem number represent...

Math 140Quiz 6 - Summer 2004

Solution Review

(Small white numbers next to problem number represent its difficulty as per cent getting it wrong.)

Problem 1 (12)

Find a general term, an: 4 , 16 , 64 , 256 , 1024.Test for arithmetic sequence:

n = 1 2 3 4

an+1 - an= 16-4= 12 64-16= 48 256-64 =192 1024-246 =768

No common difference => not arithmetic sequence

Test for geometric sequence:

n = 1 2 3 4

an+1 / an= 16/4 = 4 64/16 = 4 256/64 = 4 1024/246 = 4

Common ratio r = 4 => an = a1r(n-1) = 4(4)(n-1) = 4n

Problem 2 (19)

Find a general term, an: 0, 2, 6, 12, 20.Approach when choices are given - Test each choice:

Choice n = 1 2 3 4 5 OK?

A) an= 4n - 6 -2 2 6 10 16 NO

B) an= 2n – 2 0 2 4 6 8 NO

C) an= n2 – n 0 2 6 12 20 YES

D) an= 2(n-1) –1 0 1 3 7 15 NO

Problem 2 cont’d (19)

Find a general term, an: 0, 2, 6, 12, 20.Approach if no choices - Test for arithmetic sequence:

n = 1 2 3 4

an+1 - an= 2 – 0= 2 6 - 4= 4 12 - 6= 6 20 -12= 8



n = 1 c 3 4

an+1 / an= 2/0 = 6/4 = 3 12/6 = 2 20/12 = 5/3

No common ratio => not geometric sequence


Find a general term, an: 0, 2, 6, 12, 20.

Note pattern in arithmetic sequence test: an+1 - an = 2n.

n = 1 2 3 4

an+1 – an = 2 4 6 8

2n = 2 4 6 8

Hence, an = an -1 + 2(n - 1) = an -2 + 2(n - 1) + 2(n - 2)

2

)11)(1(21

nna

Thus, an = a1 + n(n - 1) = n2 - n.

1

11 2

nk

k

ka 2 1

11

nk

kn kaa

Problem 3 (31)

Write out the first five terms of the sequence to two decimal places. an = (-1)(n - 1) (n + 1)/(2n - 1)

Put n = 1, 2, 3, 4, 5 in the expression for an and evaluate to get:

a1 = (-1)(1 - 1) (1 + 1)/(2•1 - 1) = 2

a2 = (-1)(2 - 1) (2 + 1)/(2•2 - 1) = -3/3 = -1

a3 = (-1)(3 - 1) (3 + 1)/(2•3 - 1) = 4/5 = 0.80

a4 = (-1)(4 - 1) (4 + 1)/(2•4 - 1) = -5/7 = -0.71

a5 = (-1)(5 - 1) (5 + 1)/(2•5 - 1) = 6/9 = 0.67

Problem 4 (37) Find the first term and give a formula for the

sequence: 10th term is 16; 15th term is –29.

Note choices are of arithmetic sequence: (a1 = a) _____________ an = a + d(n - 1).

a10 = a + d(10 - 1) = 16 (a)

a15 = a + d(15 - 1) = -29 (b)

Compute: (b) minus (a) & substitute in (a) =>

d(14 - 9) = -29 –16 => d = -45/5 = -9

a - 9(9) = 16 => a = 97

an = a + d(n - 1) = 97 - 9(n - 1) = 106 - 9n

Problem 5 (12)

Express the sum using summation notation,

S = 3 + 12 + 27 + . . . + 108

Since choices are given, test each choice:

Choice S OK?

A) ak= 3k2

B) ak= 3k2

C) ak= k2

D) ak= 32k

1087548271230 6

0

k

kka NO

635261941 6

1

k

kka NO

45546327189 6

1

k

kka NO

108754827123 6

1

k

kka YES

Problem 6 (12)

Find the common difference,

an: 3.15, 4.82, 6.49, 8.16, ....

n = 1 2 3

an+1 – an = 4.82 - 3.15 = 1.67

6.49 - 4.82 = 1.67

8.16 - 6.49 = 1.67

Common difference d = 1.67

Problem 7 (31)

Find the common ratio for the geometric sequence. If a sequence is not geometric, say so.

an: 1, -3, 9, -27, 81


n = 1 2 3 4

an+1 / an= -3/1= -3 9/-3= -3 -27/9= -3 81/-27=-3

Common ratio r = -3 => an = a1r(n-1) = (-3)(n-1)

Problem 8 (81)

Find the requested sum of the arithmetic sequence.

Use the formula for the sum of first n integers:

1368

1

i

i

i

Put n = 1368 and find:

S1368 = 1368(1368 + 1)/2 = 936,396

2/)1( 1

nniSni

in

Problem 9 (81)

Find the sum, if it exists, for the infinite geometric sequence.

Factor one power of (1/5) to put into standard infinite geometric series form and use formula for the sum of an infinite geometric series with a = 2/5 & r = 1/5 < 1.

1

)5/1(2 i

i

)1/( 1

)1( raarSi

i

2/1)4/5)(5/2(

)]5/1(1/[)5/2()5/1)(5/2( 1

)1(

i

iS

Problem 10 (25)

Determine whether the given sequence is arithmetic,

geometric, or neither. If arithmetic, find the common

difference. If geometric, find the common ratio.

{an}= {5n2 - 4}

Because n is squared, {an} is not an arithmetic sequence.

Because n is not as power of a constant, rn, {an} is not a geometric sequence.

Numerical tests for arithmetic & geometric sequences are on next slide.


{an}= {5n2 - 4}Arithmetic sequence test: an+1 - an=5[(n+1)2 - n2]= 10n + 5

n = 1 2 3 4

an+1 - an= 15 25 35 45


Geometric sequence test: an+1 /an=[5(n+1)2 –4]/(5n2-4)

n = 1 2 3 4

an+1 / an= 16/1 = 16 41/16 76/41 121/76

No common ratio => not geometric sequence

Problem 11 (44)

Write the indicated term of the binomial expansion.

(3x + 2)5; 5th term

Note: in the formula for the binomial expansion,

,),( ) ( 0

ni

i

inin yainCay

xx 2403)16(5

the mth term has i = m – 1. Thus, the mth term is:.)1,( 11 mnm yamnC

Here we have n = 5, m = 5, a = 2, & y = 3x. Thus, )3(2)]!1!4/(!5[)3(2)15,5( 415515 xxC

Next slide has alternate method.

Problem 11 Cont’d (62)

Write the indicated term of the binomial expansion.

(3x + 2)5; 5th term

Here is an alternate method. Consider Pascal’s Triangle where each entry is sum of 2 above.

15101051

14641

1331

121

11

1

5th power => 5th line

Thus, term in (y + a)5 is 5a4y.

Here a = 2 & y = 3x => 5a4y = 5•24•3x = 240x

5th term => next to last entry

Problem 12 (44)

Evaluate the expression: P(8, 7).

P(n, r) = n!/(n – r)!

P(8, 7) = 8!/(8 – 7)! =

8•7•6•5•4•3•2•1!/1! = 40,320

Problem 13 (56)

Evaluate the expression: P(11, 11).

P(n, r) = n!/(n – r)!

P(11, 11) = 11!/(11 – 11)! =

11•10•9•8•7•6•5•4•3•2•1•0!/0! = 39,916,800

Problem 14 (44)

Evaluate the expression: C(9, 3).

C(n, r) = n!/[r!(n – r)!]

C(9, 3) = 9!/[3!(9 – 3)!] =

9•8•7•6!/[3•2•1•6!] = 84

Problem 15 (62) In a survey of 48 hospital patients, 15 said they were

satisfied with the nursing care, 23 said they were satisfied with the medical treatment, and 5 said they were satisfied with both. a) How many patients were satisfied with neither? b) How many were satisfied with only the medical treatment?

A: nursing satisfiedQuick solution:

B: medical satisfiedn A B n A n B n A B( ) ( ) ( ) ( )

= 15 + 23 - 5 = 3348 - 33 = 15 satisfied with neither

satisfiedmedicalonly18523)()( BAnBn

Universe is 48 people. Satisfied by in A = 15 nursing and in B = 23 medical. In AB = 5 satisfied by both nursing & medical. In AB = 15+23-5 = 33 ok with nursing or medical. In (AB) = 48-33 = 15 satisfied by neither. In B - AB = 23-5 = 18 only medical satisfied.

A= 15 B= 23

(AB)=15

AB=5


Problem 16 (69)

How many 4-letter codes can be formed using the letters A, B, C, D, E, and F? No letter can be used more than once.The first code position can be filled in 6 ways, the second in 5, the third in 4, & the fourth in 3.

Thus, by the Multiplication Rule there are:

6•5•4•3 = 360 ways.

Or, by the Permutation Rule for 6 objects taken 4 at a time, there are:

P(n, r) = n!/(n – r)! = 6!/(6–4)! =

6•5•4•3•2! /2! = 6•5•4•3 = 360 ways.

Problem 17 (56)

In how many ways can 8 volunteers be assigned to 8 booths for a charity bazaar?

The first booth can be staffed in 8 ways, the second in 7, the third in 6,....

By the Multiplication Rule there are:

8•7•6•...1 = 8! ways.

That is, 8•7•6•5•4•3•2•1 = 40,320 ways.

Problem 18 (37)

Tell whether or not the given model is a probability model.

The probabilities in Model I all are in [0, 1] and add to 1. So YES, it is a probability model.

The probabilities in Model II all are in [0, 1] but add to 1.27 > 1. So NO, it is not a probability model.

Problem 19 (31) A bag contains 7 red marbles, 3 blue marbles, and 1

green marble. What is the probability of choosing a marble that is not blue when one marble is drawn from the bag?

There are 7 + 3 + 1 = 11 marbles in the bag. Picking a “not blue” one means picking a red or green one of which there are 8.

The probability is, thus, 8/11.

Problem 20 (25)

Two 6-sided dice are rolled. What is the probability that the sum of the two numbers on the dice will be greater than 10?

There are 6•6 = 36 ways that the dice may fall.

To get a sum > 10 one must have a sum of 11 or 12.

There are only 3 ways to do this: 5+6, 6+5, or 6+6.

The probability is, thus, 3/36 = 1/12.

Problem 21 (75)

A hamburger shop sells hamburgers with cheese, relish, lettuce, tomato, onion, mustard, or ketchup. How many different hamburgers can be concocted using any 3 of the extras?

Since order is unimportant, we seek the number of combinations of 7 dressings taken 3 at a time.

By the Combination Rule for 7 objects taken 3 at a time, there are:

C(n, r) = n!/[r!(n – r)!] = 7!/[3!(7–3)!] =

7•6•5•4! /(3•2•1•4!) = 35 ways.

math 140 quiz 6 - summer 2004 solution review (small white numbers next to problem number represent...

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