course Mth 151 008. Arithmetic Sequences
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So all I have to do is divide the amount of numbers and see what a pair adds up to it then times the pairs to the total of a pair ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since there are 2,000 numbers that means there is 1,000 pairs 1+2,000= 2,001 2+1,999=2,001 So next we would times the 1,000 * 2,001= 2001000 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we have 501 that means we have 250.5 pairs Next 501+1= 502 500+2= 502 502 * 250.5 = 12751 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1533 has 766.5 pairs 1533+1= 1534 1532+2= 1534 1534 * 766.5= 1175811 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Sorry I have to comment on the (you've got a calculator, and I've only got my unreliable head) that made me laugh so hard!!
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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): How you get the sums make a lot of since now!!! I got really confused and I didn’t understand why you added one until you showed the students comment basically all I have to do is subtract the number that comes before the given number. In this case the number was 55 so I would have to subtract 54. In the one above the first number was 5 so you subtract 4 from 15 to get the pairs. ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First find out how many actual numbers you have 900-3= 897 So you take 897/2 to find out the pairs 897/2= 448.5 Next find out what the sum of two pairs will be 4+900= 9004 5+899= 904 Now you can find out how much the sum really is 904 * 448.5 = 405444 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): WAIT why would you divide by 4 to find out how many sums you get, I though you would divide by 2 since there is only 2 numbers in a pair. Now I really am confused
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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* "