course mth 151
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.
008. Arithmetic Sequences
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q001. There are seven questions in this set.
See if you can figure out a strategy for quickly adding the numbers 1 + 2 + 3 + ... + 100, and give your result if you are successful. Don't spend more than a few minutes on your attempt.
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Your solution: I can? do it fast math is not my strong suit. I give up it has already been 10 minutes and I am no closer to solving than when I first started.
10 minutes is about right.
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Given Solution:
These numbers can be paired as follows:
1 with 100,
2 with 99,
3 with 98, etc..
There are 100 number so there are clearly 50 pairs. Each pair adds up to the same thing, 101. So there are 50 pairs each adding up to 101. The resulting sum is therefore
total = 50 * 101 = 5050.
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Self-critique (if necessary):
I see it is easier than I was making it.
Self-critique Rating:
Question: **** `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000.
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Your solution:
So if like before I can pair up 1 and 2000 to = 2001, 2 to 1999=2001, and 3 with 1998 equaling 2001. There are 1000 pairs = to 2001 so I would multiply 1000 by 2001 to get 2001000.
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Given Solution:
Pairing 1 with 2000, 2 with 1999, 3 with 1998, etc., and noting that there are 2000 numbers we see that there are 1000 pairs each adding up to 2001.
So the sum is 1000 * 2001 = 2,001,000.
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Self-critique (if necessary):
That is easy once you know what you are doing.
Self-critique Rating:
Question: **** `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501.
This told the answer before I could try it. Was this purposeful?
We can pair 1 with 501, 2 with 500, 3 with 499, etc., and each pair will have up to 502. However there are 501 numbers, so not all of the numbers can be paired. The number in the 'middle' will be left out.
However it is easy enough to figure out what that number is, since it has to be halfway between 1 and 501. The number must be the average of 1 and 501, or (1 + 501) / 2 = 502 / 2 = 266. Since the other 500 numbers are all paired, we have 250 pairs each adding up to 502, plus 266 left over in the middle.
The sum is 250 * 502 + 266 = 125,500 + 266 = 125,751.
Note that the 266 is half of 502, so it's half of a pair, and that we could therefore say that we effectively have 250 pairs and 1/2 pair, or 250.5 pairs.
250.5 is half of 501, so we can still calculate the number of pairs by dividing the total number of number, 501, by 2.
The total sum is then found by multiplying this number of pairs by the sum 502 of each pair:
250.5 * 502 = 125,766.
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Self-critique (if necessary):
How is 266 half of 502 when I get 251?
You're right.
Self-critique Rating:
Question: **** `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533.
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Your solution: I divided 1533 by 2 and got 766.5 and multiplied that by 1533 and my answer is 1175044.5
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Given Solution:
Pairing the numbers, 1 with 1533, 2 with 1532, etc., we get pairs which each adult to 1534. There are 1533 numbers so there are 1533 / 2 = 766.5 pairs. We thus have a total of 1534 * 766.5, whatever that multiplies out to (you've got a calculator, and I've only got my unreliable head).
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Self-critique (if necessary):
That? how I did it.
Self-critique Rating:
Question: **** `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945.
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Your solution:
I divide 945 by 2 and get 472.5 and I then multiply that by 945 and get 446512.5
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Given Solution:
We can pair up 55 and 945, 56 and 944, etc., obtaining 1000 for each pair. There are 945 - 55 + 1 = 891 numbers in the sum (we have to add 1 because 945 - 55 = 890 tells us how many 1-unit 'jumps' there are between 55 and 945--from 55 to 56, from 56 to 57, etc.. The first 'jump' ends up at 56 and the last 'jump' ends up at 945, so every number except 55 is the end of one of the 890 'jumps'. But 55 is included in the numbers to be summed, so we have 890 + 1 = 891 numbers in the sum).
If we have 891 numbers in the sum, we have 891/2 = 445.5 pairs, each adding up to 1000.
So we have a total of 445.5 * 1000 = 445,500.
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Self-critique (if necessary):
I didn? start at 55 if I did I would have subtracted 55 from 945 to get 890 dividing that by 2 getting 445 so 445 * 890 is 396050. How did we get 1000 pairs from this?
Self-critique Rating:
Question: **** `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900.
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Your solution:
I divided 900 by 4 because each number goes up by 4 getting 225 pairs. So 225 *900 = 202500.
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Given Solution:
Pairing 4 with 900, 8 with 896, etc., we get pairs adding up to 904. The difference between 4 and 900 is 896.
The numbers 'jump' by 4, so there are 896 / 4 = 224 'jumps'. None of these 'jumps' ends at the first number so there are 224 + 1 = 225 numbers.
Thus we have 225 / 2 = 112.5 pairs each adding up to 904, and our total is 112.5 * 904.
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Self-critique (if necessary):
I keep forgetting to add I do this on scrap paper but then when I go to type it I forget.
Self-critique Rating:
Question: **** `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number?
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Your solution:
n(n+1) divided by 2
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Given Solution:
We can pair 1 and n, 2 and n-1, 3 and n-2, etc., in each case obtaining a sum of n + 1. There are n numbers so there are n/2 pairs, each totaling n + 1. Thus the total is n/2 * (n+1).
I thought you wanted the formula.
The formula is OK, but what I really want to see is the reasoning behind the formula.
&#Good responses. See my notes and let me know if you have questions. &#