#$&* course MTH 151 10/6/14 around 10:30 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: These numbers can be paired as follows: 1 with 100, 2 with 99, 3 with 98, etc.. There are 100 number so there are clearly 50 pairs. Each pair adds up to the same thing, 101. So there are 50 pairs each adding up to 101. The resulting sum is therefore total = 50 * 101 = 5050. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q002. See if you can use a similar strategy to add up the numbers 1 + 2 + ... + 2000. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1000 different pairs, each adding up to be 2001 1+2000 2+1999 3+1998,... 1000*2001=2,001,000 total sum again we could use the equation T_2000=1/2*2000*(2000+1) = 1000*2001 = 2,001,000 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q003. See if you can devise a strategy to add up the numbers 1 + 2 + ... + 501. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we can have 250 pairs in this with each pair adding up to 502 1+500 2+499 3+498,... however we end up with one number being left out of the pair, this number ends up being the one in the middle, 251 so we can take our regular equation and just add this extra number to it 250*502+251=125,500+251= 125,751 sum total if we use the equation T_501=1/2*501*(501+1)=250.5*502=125,751 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We can pair 1 with 501, 2 with 500, 3 with 499, etc., and each pair will have up to 502. However there are 501 numbers, so not all of the numbers can be paired. The number in the 'middle' will be left out. However it is easy enough to figure out what that number is, since it has to be halfway between 1 and 501. The number must be the average of 1 and 501, or (1 + 501) / 2 = 502 / 2 = 251. Since the other 500 numbers are all paired, we have 250 pairs each adding up to 502, plus 251 left over in the middle. The sum is 250 * 502 + 251 = 125,500 + 251 = 125,751. Note that the 251 is half of 502, so it's half of a pair, and that we could therefore say that we effectively have 250 pairs and 1/2 pair, or 250.5 pairs. 250.5 is half of 501, so we can still calculate the number of pairs by dividing the total number of number, 501, by 2. The total sum is then found by multiplying this number of pairs by the sum 502 of each pair: 250.5 * 502 = 125,766. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q004. Use this strategy to add the numbers 1 + 2 + ... + 1533. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1533/2= 766 whole pairs or 766.5 1+1533 2+1532 3+1531,... all pairs add up to 1534 middle number is 764 766*1534+767 = 1,175,811 using the equation T_1533=1/2*1533*(1533+1)=766.5*1534 = 1,175,811 confidence rating #$&*: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q005. Use a similar strategy to add the numbers 55 + 56 + ... + 945. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 945-54=891 total numbers you only subtract 54 because there are only 54 counting numbers not there (1,2,3,...,53,54) 55 is still in this set 445 total pairs 55+945 56+944 57+943,... each pair adds up to 1000 445*1000= 445,000 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We can pair up 55 and 945, 56 and 944, etc., obtaining 1000 for each pair. There are 945 - 55 + 1 = 891 numbers in the sum (we have to add 1 because 945 - 55 = 890 tells us how many 1-unit 'jumps' there are between 55 and 945--from 55 to 56, from 56 to 57, etc.. The first 'jump' ends up at 56 and the last 'jump' ends up at 945, so every number except 55 is the end of one of the 890 'jumps'. But 55 is included in the numbers to be summed, so we have 890 + 1 = 891 numbers in the sum). If we have 891 numbers in the sum, we have 891/2 = 445.5 pairs, each adding up to 1000. So we have a total of 445.5 * 1000 = 445,500. STUDENT COMMENT I got very confused on this one. I don’t quite understand why you add a 1. INSTRUCTOR RESPONSE For example, how many numbers are there in the sum 5 + 6 + 7 + ... + 13 + 14 + 15? 15 - 5 = 10. However there are 11 numbers in the sum (5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q006. Devise a strategy to add the numbers 4 + 8 + 12 + 16 + ... + 900. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4+900 8+896 12+892,... 900/4=225 total numbers we divide by 4 because it jumps by 4 each time each pair adds up to 904 225/2= 112.5 pairs 112.5*904= 101,700 confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q007. What expression would stand for the sum 1 + 2 + 3 + ... + n, where n is some whole number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we can pair 1+n, 2+(n-1), 3+(n-2),... each pair gives you a sum of n+1 and there are n/2 pairs the total sum is n/2*(n+1) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We can pair 1 and n, 2 and n-1, 3 and n-2, etc., in each case obtaining a sum of n + 1. There are n numbers so there are n/2 pairs, each totaling n + 1. Thus the total is n/2 * (n+1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q008. What are the following two sums? •50 + 51 + 52 + ... + 998 + 999 + 1000 •3 + 6 + 9 + ... + 300 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1000-49=951 total numbers we can pair 50+1000 51+999 52+998,... 951/2= 475.5 pairs each pair adds up to 1050 475.5*1050 = 499,275 300/3=100 total numbers we divide by 3 because the numbers jump by 3 each time 100/2=50 pairs 3+300 6+297 9+294,... each pair adds up to 303 50*303= 15,150 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!