math13open

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course Mth 151

2/14/2012 7:25 PM

008. Arithmetic Sequences 

 

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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:

If you add the first and last number, then the second and second to last number and keep going you will have 50 pairs of 101 and if you multiply them together 5050. 

 

 

confidence rating #$&*: 3

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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): OK

 

 

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Self-critique Rating: Ok

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Question: `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000.

 

 

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Your solution:

If you do the same as before we get 100 sets of 2001 and if you multiply 1000 times 2001 you get 2001000.

 

 

confidence rating #$&*:

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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):

 

 

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Self-critique Rating:

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Question: `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501.

 

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Your solution:

Doing the same as before I got 250 pairs of 502 and I multiplied them together and got 125500. But since there is an odd number of numbers were adding we need to add the middle number because it was left out. The middle is 251 and if you add that to 125500 you get 125751.  

 

 

confidence rating #$&*: 3

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Given Solution:

 

 

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 = 251. Since the other 500 numbers are all paired, we have 250 pairs each adding up to 502, plus 251 left over in the middle.

 

The sum is 250 * 502 + 251 = 125,500 + 251 = 125,751.

 

Note that the 251 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): OK

 

 

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Self-critique Rating: OK

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Question: `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533.

 

 

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Your solution:

First I got half of 1533 which is 766.5, then I multiplied that by 1534, because that is what the sum of each pair equals.

776.5 * 1534 = 1,175,811

 

 

confidence rating #$&*: 3

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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): OK

 

 

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Self-critique Rating: OK

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Question: `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945.

 

 

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Your solution:

First I found out how many numbers I was adding together by subtracting 54 from 945 which equaled 891. Then I got half of 891 which is 445.5, then I multiplied that by 1000, because that is what the sum of each pair equals.

445.5 * 1000 = 445,500

 

 

confidence rating #$&*: 3

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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.

 

STUDENT COMMENT

 

I got very confused on this one. I don’t quite understand why you add a 1. 

INSTRUCTOR RESPONSE

 

For example, how many numbers are there in the sum 5 + 6 + 7 + ... + 13 + 14 + 15?

15 - 5 = 10.

However there are 11 numbers in the sum (5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15).

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question: `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900.

 

 

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Your solution:

Pairing the numbers together I got 904 every time. Then I divided 900 by 4 to figure how many numbers there are. There 225 numbers so I divided that by 2 to get 112.5 pairs. Then I multiplied the number of pairs by 904 to get 101,700

 

 

confidence rating #$&*: 3

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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): OK

 

 

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Self-critique Rating: OK

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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 would also be the number of numbers your adding, so to get the number of pairs you would divide n by 2. Then I would multiply that by n + 1 because that is what the sum of each pair would be. So the expression would be 2/n * (n + 1)

 

 

confidence rating #$&*: 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).

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question: `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number?

 

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

N would also be the number of numbers your adding, so to get the number of pairs you would divide n by 2. Then I would multiply that by n + 1 because that is what the sum of each pair would be. So the expression would be 2/n * (n + 1)

 

 

confidence rating #$&*: 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).

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question: `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number?

 

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

N would also be the number of numbers your adding, so to get the number of pairs you would divide n by 2. Then I would multiply that by n + 1 because that is what the sum of each pair would be. So the expression would be 2/n * (n + 1)

 

 

confidence rating #$&*: 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).

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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