Query 7

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course Mth 151

You may want to double-check the link to your video in section 1.2 called Successive Differences. It appears to have nothing to do with what we are studying.

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.

007. `Query 7

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Question: `qQuery 1.2.18 (previously 1.2.6) seq 2, 57, 220, 575, 1230, 2317 ... by successive differences

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3992

2, 57, 220, 575, 1230, 2317, 3992

55 163 355 655 1087 1675

108 192 300 432 588

84 108 132 156

24 24 24

confidence rating #$&*: 3

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Given Solution:

`a** Using sequences of differences we obtain:

2, 57, 220, 575, 1230, 2317, # 3992

55, 163, 355, 655, 1087, # 1675

108, 192, 300, 432, # 588

84, 108, 132, # 156

24, 24,

The final results, after the # signs, are obtained by adding the number in the row just below, in the following order:

Line (4) becomes 132+24=156

Line (3) becomes 432+156=588

Line (2) becomes 1087+588=1675

Line (1) becomes 2317+1675=3992

The next term is 3992. **

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `q 1.2.30 (previously 1.2.18) 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc.

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Your solution:

1˄2 + 1 = 2˄2 - 2 1 + 1 = 4 - 2 2 = 2 True

2˄2 + 2 = 3˄2 - 3 4 + 2 = 9 - 3 6 = 6 True

3˄2 + 3 = 4˄2 - 4 9 + 3 = 16 - 4 12 = 12 True

The next equation would be:

4˄2 + 4 = 5˄2 - 5 16 + 4 = 25 - 5 20 = 20 True

confidence rating #$&*: 3

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Given Solution:

`a** The next equation in the sequence would be 4^2 + 4 = 5^2 - 5

The verification is as follows:

4^2 + 4 = 5^2 - 5 simplifies to give you

16 + 4 = 25 - 5 or

20 = 20 **

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `q1.2.42 (previously 1.2.30) state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3

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Your solution:

This is a special sum formula.

The left side of this equation is the square of the sum of all the numbers 1 through n.

The right side of this equation is the sum of all the numbers cubed.

The left side of the equation is equal to the right side of the equation

confidence rating #$&*: 2

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Given Solution:

`a** the equation says that the square of the sum of the first n counting numbers is equal to the sum of their cubes **

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `q (previously 1.2.36)

Divide the first triangular number by 3 and write down the remainder.

Divide the second triangular number by 3 and write down the remainder.

Divide the third triangular number by 3 and write down the remainder.

Divide the fourth triangular number by 3 and write down the remainder.

Continue until you have established a pattern.

What is the pattern?

Can you explain why the pattern occurs? Is it possible that the pattern doesn't go on forever?

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Your solution:

The pattern is that the remainders are 1, 0, 0, 1, 0, 0, 1, 0, 0,

This pattern occurs because every third number is not evenly divisible by 3, which gives it a remainder of 1. Every 2nd and 3rd number is evenly divisible by 3.

It is possible that this pattern could go on forever. This is inductive reasoning so may be valid, but no necessarily true.

confidence rating #$&*: 3

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Given Solution:

`a** The triangular numbers are 1, 3, 6, 10, 15, 21, . . . . We divide these by 3 and get the sequence of remainders.

When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0.

It turns out that the sequence continues as a string of 1,0,0 's. At this point that is an inductive pattern, but remmeber that the sequence of triangular numbers continues by adding successively larger and larger numbers to the members of the sequence. Since the number added always increases by 1, and since every third number added is a multiple of 3, is isn't too difficult to see how the sequence of remainders comes about and to see why it continues as it does.

COMMON ERROR: .3333333,1,2,3.3333333,etc.

INSTRUCTOR CORRECTION:

You need the remainders, not the decimal equivalents. When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's.

COMMON ERROR: 1/3, 1, 2, 3 1/3

CORRECTION:

These are the quotients. You need the remainders. If you get 1/3 that means the remainder is 1; same if you get 3 1/3. If you just getting whole number (like 1 or 2 in your calculations) the remainder is 0.

In other words, when you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1.

The remainders form a sequence 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's. **

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `q 1.2. 58 (previously 1.2.48) use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The formula for finding Octagonal numbers is:

On = n(6n - 4) / 2

O12 = 12(6x12-4) / 2

O12 = 12 x 68 / 2

O12 = 816 / 2

O12 = 408

confidence rating #$&*: 2

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Given Solution:

`a** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows (you should know this pattern; you might need one or more of these formulas on the test):

Triangular numbers: n / 2 * [ n + 1 ] note that this is the same as Gauss' formula

Square numbers: n / 2 * [ 2n + 0 ] or just n^2

Pentagonal #'s: n / 2 * [ 3n - 1 ]

Hexagonal #'s: n / 2 * [ 4n - 2 ]

Heptagonal #'s: n / 2 * [ 5n - 3 ]

Octagonal #'s: n / 2 * [ 6n - 4 ]

The coefficient of n in the bracketed term starts with 1 and increases by 1 each time, and the +1 in the first bracketed term decreases by 1 each time.

You will need to know these formulas for the test.

The last formula is for octagonal numbers. To get n = 12 octangonal number use n/2 * [ 6n - 4 ] to get 12 / 2 * [ 6 * 12 - 4 ] = 6 * 68 = 408. **

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q 1.2. 58 (previously 1.2.48) use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The formula for finding Octagonal numbers is:

On = n(6n - 4) / 2

O12 = 12(6x12-4) / 2

O12 = 12 x 68 / 2

O12 = 816 / 2

O12 = 408

confidence rating #$&*: 2

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Given Solution:

`a** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows (you should know this pattern; you might need one or more of these formulas on the test):

Triangular numbers: n / 2 * [ n + 1 ] note that this is the same as Gauss' formula

Square numbers: n / 2 * [ 2n + 0 ] or just n^2

Pentagonal #'s: n / 2 * [ 3n - 1 ]

Hexagonal #'s: n / 2 * [ 4n - 2 ]

Heptagonal #'s: n / 2 * [ 5n - 3 ]

Octagonal #'s: n / 2 * [ 6n - 4 ]

The coefficient of n in the bracketed term starts with 1 and increases by 1 each time, and the +1 in the first bracketed term decreases by 1 each time.

You will need to know these formulas for the test.

The last formula is for octagonal numbers. To get n = 12 octangonal number use n/2 * [ 6n - 4 ] to get 12 / 2 * [ 6 * 12 - 4 ] = 6 * 68 = 408. **

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Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Good work. Let me know if you have questions. &#