If your solution to stated problem does not match the given solution, you should self-critique per instructions at

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

SOLUTIONS/COMMENTARY ON QUERY 17

  ****   Query   11.3.26  a = -2, r = 4.

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                                                  11:39:02

Your solution:

Confidence Assessment:

Given Solution:

 ** Since a(n) = a * r^(n-1) we have a(n) = -2 * 4^(n-1).

Thus a(5) = -2 * 4^(5-1) = -2 * 4^4 = -512. **

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                                                  11:39:03

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  ****   Give the 5th term and the nth term of the geometric sequence.

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                                                  11:40:51

Your solution:

Confidence Assessment:

Given Solution:

STUDENT ANSWER      

 we can find the 5th term using the same formula a sub 5=(-2)(4)^5-1

a sub 5=(-2)(256)

a sub 5= (-512)

a sub n=(-2)(4)^(n-1)

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                                                  11:40:52

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  ****   Query   11.3.42  sum (4 * 3^(k-1), k, 1, 4)

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                                                  11:51:36

Your solution:

Confidence Assessment:

Given Solution:

 first we can use the formula

S sub n= a(1-r^n/1-r)

S sub n=4(1-3^n)/1-3)

S sub n=4-4(3)^n/ -2

-2 S sub n=4-12n  ** 4 * 3^n is not 12 * n.  It's not 12^n either.  It's just 4 * 3^n. **

S sub n=6n-2

** If the sum is from k = 1 to 4 then you are adding 4 * 3^0 + 4 * 3^1 + ... + 4 * 3^3.

This is the sum of the first 4 terms of the sequence.

Note that S(n) = a * (1 - r)^n / (1 - r). 

This sum will be S4 = 4 * (1 - 3^4) / (1 - 3) = 4 * (1 - 81) / (1 - 3) = 4 * 40 = 160. **

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  ****   Query   11.3.66 (5th edition 70) ball dropped from 30 feet, .8 of

height on each bounce.

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                                                  12:11:52

Your solution:

Confidence Assessment:

Given Solution:

 a) after the third bounce it will reach a height of 15.36 ft.

b) after the nth bounce it will be at a height of

S sub n=30((1-(0.8^n)/(1-0.8)

c) after the 19th bounce its height will be less than 6 inches.

d)it will travel 149.639 ft.

**After the 1st bounce it will bounce back up to .8 * 30 ft; after the 2d bounce the height will be .8^2 * 30 ft; after the 3d bounce height is .8^3 * 30 ft.

After the nth bounce the height is .8^n * 30 ft.

The ball falls 30 feet, then it bounces to height 24 feet and back, then to height 19.2 feet, etc.; the distance it travels is 30 + 48 + 38.4 + etc..

In symbols the distance is 30 + 2 * ( .8 * 30 + .8^2 * 30 + .8^3 * 30 + ... + .8^19 * 30), which upon factoring out .8 * 30 from the second part becomes

30 + 2 * (.8 * 30) ( 1 + .8 + .8^2 + ... + .8^18), or

30 + 48 * ( 1 + .8 + .8^2 + ... + .8^18).

The sum ( 1 + .8 + .8^2 + ... + .8^18) is, by the formula, equal to

a * (1 - r^18) / (1 - r) = 1 * (1 - .8^18) / (1 - .8) = 4.91, approximately.

So the total distance through the 18th bounce is

30 + 48 * 4.91 = 265.7, approximately. 

Adapting the sum 30 + 48 * ( 1 + .8 + .8^2 + ... + .8^18) to the total of all bounces we get 30 + 48 * ( 1 + .8 + .8^2 . . . + .8^n + . . . ). 

The sum 1 + .8 + .8^2 + . . . + .8^n + . . . is 1 / (1-r) = 1 / (1-.8) = 5.  So the total distance on all bounces is

total distance = 30 + 48 * 5 = 270. **

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                                                  12:11:52

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  ****   To what height will the ball bounce after the third time it hits the ground and how did you obtain your answer?

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                                                  12:12:18

Your solution:

Confidence Assessment:

Given Solution:

19.2(0.8)=15.36 ft.

 ** After the 1st bounce it will bounce back up to .8 * 30 ft; after the 2d bounce the height will be .8^2 * 30 ft; after the 3d bounce height is .8^3 * 30 ft.

After the nth bounce the height is .8^n * 30 ft.

After the 3d bounce the height is .8^3 * 30 ft = 15.36 ft. **

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                                                  12:12:18

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  ****   To what height will the ball bounce after the nth time it hits the ground and how did you obtain your answer?

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                                                  12:13:01

Your solution:

Confidence Assessment:

Given Solution:

 S sub n=30((1-(0.8 ^n)/(1-0.8)

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                                                  12:13:02

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  ****   After how many bounces will ball's height first be less than six

inches?

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                                                  12:23:53

Your solution:

Confidence Assessment:

Given Solution:

 ** Solve for bounce n when height is 6 in = . 5 ft:

30 * .8^n = .5

.8^n = .5 / 30

.8^n = .01666

n = log(.01666) / log(.8) = 18.3 approx.

So it will first bounce less than 6 inches on the 19th bounce. **

on the 8th bounce.

I worked the equation ar^(n-1) for a(n) = 6, and solved for n

 30(.8)^(n-1) = 6

.8^(n-1) = 6/30  ( reduced to 1/5)

n - 1 = 'log .8 (1/5)

n = 1 + ['ln (1/5) / 'ln(.8) ]

n = 7.213

So th last 6 inch height will occur on the 7th bounce, on the 8th bounce the ball will bounce less than 6 inches.

** Right method, good solution, except for two things.  The height function is 30 * .8^n, not 30 * .8^(n-1). 

But your equation sets height equal to 6, which means 6 feet, not 6 inches.

Your function gives heights in feet.  At 6 inches the height is .5 ft.  You would solve

30 * .8^n = .5. **

19th bounce

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                                                  12:23:54

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  ****   Query   11.5.6 evaluate the expression C(100,98).

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                                                  12:42:26

Your solution:

Confidence Assessment:

Given Solution:

STUDENT ANSWER:

we first use the formula which states n taken j at a time is (n/j)=

n!/j!(n-j)!

so 100!/98!2! so we have (9.3326 * 10^157)/1.96*10^155 so we have 476.15

** Never evaluate the factorials then divide.

 Calculator solutions to this problem are unacceptable, except that the calculator can be used for the very last step where the integers that don't divide our are multiplied.  The expressions must be written out and evaluated in a manner similar to the following:

100! / (98! * 2!) = 100 * 99 * 98 * 97 * .... * 1 / [( 98 * 97 * ... * 1) * (2 * 1)].

The 98 * 97 ... in the denominator divides out the same mess in the numerator leaving

100 * 99 / ( 2 * 1) = 50 * 99 = 4950.

Combinations and permutations never result in fractional answers.  The denominators ALWAYS divide out completely.

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  ****   Query   11.5.18 expand (2x+3)^5.

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                                                  12:50:20

Your solution:

Confidence Assessment:

Given Solution:

STUDENT SOLUTION

first we can use the binomial theorem  and set the equation like this

(5,0)2x^5+(5,1)6x^4+(5,2)18x^3+(5,3)54x^2+(5,4)81x+(5,5)243

so we have

2x^5+720x^4+360x^3+540x^2+405x+243

I used the Binomial Theorem

SUM (from j = o to n) of the expression (n,j) a^jx^(n-j)

to expand the expression by letting x = 2x and a = 3 and n = 5

The binomial coeffiecients (n, j) with j being equal to the number of the term beginning with zero.

**(2x + 3)^5 = C (5,0) (2x)^5 + C (5,1) 3 (2x)^4 + C (5,2) 3^2 ((2x)^3 + C (5,3) 3^3 (2x)^2 + C (5,4) 3^4 (2x)^1 + C (5,5) 3^5.

Simplify this.  You get

32·x^5 + 240·x^4 + 720·x^3 + 1080·x^2 + 810·x + 243  **

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                                                  12:50:20

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  ****   Query   11.5.32 third term in expansion of (x-3)^7.

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                                                  12:52:23

Your solution:

Confidence Assessment:

Given Solution:

using the binomial theorem we find that the third term value is (7,3)27x^4

so the value is 945x^4

 ** The term would be C (7,3) * x^3 * (-3)^3 = ... = -945 x^4. 

Be careful--don't skip steps.  **

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                                                  12:52:28

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