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Deducing sums such as 1+2+3+ .+100

  1. Nov 7, 2014 #1
    1. The problem statement, all variables and given/known data
    You are given the following two sums:

    1+2+3+...+50 = 1275 and
    1+2+3+...+100 = 5050

    Using these results, find the following sums. (no long methods allowed)

    a) 51+52+53+...+100
    b) 2+4+6+...+100
    c) 1+3+5+...+99
    d) 1-2+3-4+5-...+99-100
    e) 100.01-100.02+100.03-100.04+... -101

    2. The attempt at a solution

    a) (1+2+3+...+100) - (1+2+3+...+50) = 5050 - 1275 = 3775
    b) (1+2+3+...+50) + (1+2+3+...+50) = 1275 +1275 = 2550
    c) stuck, need help on this one
    d) (-1)+(-1)...+(-1) = -100
    e) (-0.01)+(-0.01)+...+(-0.01) = 1

    Could you please check a, b, d and e and help me with c? Thank you :)
     
  2. jcsd
  3. Nov 7, 2014 #2

    PeroK

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    c) Think about c relative to b.

    d) & e) are wrong.
     
  4. Nov 7, 2014 #3
    c) Is it 2550 - 100 = 2450

    d) and e) can someone help me a little to start?
     
  5. Nov 7, 2014 #4
    Is e) -0.01)+(-0.01)+...+(-0.01) = -1
     
  6. Nov 7, 2014 #5

    PeroK

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    I think you've got c, d & e wrong for the same reason. You are not counting carefully enough.
     
  7. Nov 7, 2014 #6

    rcgldr

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    How many numbers are being added in c?

    Once you figure out c, then similar to before, think about d relative to c and b.
     
  8. Nov 7, 2014 #7
    C) is it 25 x 100 + 50 = 2550 ?
     
  9. Nov 7, 2014 #8
    Wait no c) 25 x 100 = 2500
     
  10. Nov 7, 2014 #9
    So d) would be 2500 - 2550 = -50
     
  11. Nov 7, 2014 #10
    really stuck on e)
     
  12. Nov 7, 2014 #11
    are c and d correct now and could someone guide me for e?
     
  13. Nov 7, 2014 #12

    rcgldr

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    Think about e relative to d.
     
  14. Nov 7, 2014 #13

    DaveC426913

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    Alternately, consider breaking the numbers (such as 100.01) into two components on either side of the decimal, see if you can find a calculation that efficiently sums each part, then add the two pieces back together.
     
  15. Nov 7, 2014 #14
    I think the "formula" is to bump up the last number by 1 and multiply it by half of the last number.
    Ex: 1+2+3+4
    First: 4 --> 5
    Second: 4/2= 2
    Finally 2*5=10
    This method works because my tutor got angry and made me add up all the number between 1 and 100 by hand and I did it this way. You will probably have to alter the formula on e.
     
  16. Nov 8, 2014 #15
    yes, although I'm not sure how you got this. For the requested "short method" you should probably look at the terms in (b)+(c)
    yes - can you see now what was wrong with your previous attempt for (d)? PeroK had some words you should take to heart...

    if you can work that out, you should have no problem with (e) - but again, "short method", you should be able to find (d) embedded inside (e).
     
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