query 7

course mth 151

??y?d???{???assignment #007

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Liberal Arts Mathematics I

06-13-2006

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20:45:18

Query 1.2.6 seq 2, 51, 220, 575, 1230, 2317 ... by successive differences

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RESPONSE -->

3992

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20:45:27

** If the sequence is 2, 57, 220, 575, 1230, 2317, ... then we have:

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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20:46:31

1.2.18 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc.

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RESPONSE -->

4^2+4=5^-5

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20:46:41

** 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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21:11:24

1.2.30 state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3

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RESPONSE -->

basically you are cubing each number

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21:11:31

** 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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RESPONSE -->

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions.

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21:23:56

1.2.36 1 st triangular # div by 3, remainder; then 2d etc. Pattern.

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RESPONSE -->

1/3

10/3

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21:24:22

** 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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RESPONSE -->

You were on the right track.

This also requires a self-critique.

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21:35:10

1.2.48 use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number.

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RESPONSE -->

408

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21:35:16

** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows:

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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RESPONSE -->

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You appear to understand what you are doing here. You could use a little more detail in your answers, and more consistency in self-critiquing your solutions.