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Spivak Calculus Summation problem

  1. May 6, 2015 #1
    • Member warned about not using the homework template
    Hi, I've enclosed my problem and attempt at solution below. I had problems with the latex so I photographed a picture of my work. The first top half is my attempt at the solution and the bottom is the solution that Spivak provides.

    I dont understand how he reached his solution and I dont understand how to manipulate these equations "1+....+n"

    Your feedback is much appreciated.

    Attached Files:

  2. jcsd
  3. May 6, 2015 #2
    Do you understand how he got from the first to the second line?
  4. May 6, 2015 #3
    for 1+.....+n, let's take n=11 as an example: If you think of 1+2+3+....+11, that is = (11+1) + (10+2) + (9+3) + ..... + (6+6) and these terms are all 12 or (n+1). Now, how many these (n+1) terms do you get?
  5. May 6, 2015 #4
    Once you have figured that out, then look at Spivak's first line: 1 + 3 + 5 + 7 + ....+ (2n - 1)
    What can you add to this to make it look like 1 + 2 + 3 + 4 + .......+ (2n - 1) + 2n which you know how to solve?
  6. May 7, 2015 #5
    You get n amount of n+1? Im sorry I still dont get it
  7. May 7, 2015 #6
    I add even numbers
  8. May 7, 2015 #7
    Ok maybe I should have explained a bit more what I mean and taken n to be an even number. So imagine you have n numbers where n is even, say n = 10 and you're going to add them all up together. So they're all lined up one after the next 1 2 3 4 5 6 7 8 9 10. Now you could fold this line (or wrap it) so that the 10 is under the 1, the 9 is under the 2, the 8 is under the 3 etc, like:
    1 2 3 4 5
    10 9 8 7 6

    The point of pairing them this way is because as go to the right in the top line you go +1 whereas as you go to the right in the bottom line you go -1 so that the pairs always add to 11. So now how many pairs of 11 do you get?
  9. May 7, 2015 #8
    Yes precisely you add even numbers to them and now you sum 1+2+3+...+2n and because you added all those even numbers you have to subtract them also (so that you have effectively added 0 to your equation). How do you think Spivak's term -2(1+2+...+n) is related to what I just said?
    Does it make sense?
  10. May 7, 2015 #9
    What problem did you have ?
  11. May 7, 2015 #10


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    Don't do too much of the solution for the OP.
  12. May 7, 2015 #11
    Duly noted.
  13. May 8, 2015 #12
    So then I have 1+2+3+4+5+(2n-1)+2n

    But when I subtract the sum of 2n which is 2(1+...+n)
    the second line is 1+2+3+4+5+2n-2(1+...+n), the odd number (2n-1) somehow disappeared and the number im supposed to add to get zero -2(1+...+n) is still there

    I dont understand
  14. May 8, 2015 #13


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    What do you mean by "the sum of 2n" ?
  15. May 8, 2015 #14
    I put what I meant, which is 2(1+.....+n)
  16. May 8, 2015 #15


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    Rather than calling this " the sum of 2n " it's much clearer to say something like, " the sum of all even integers from 2 through 2n. "

    In Latex that's ##\displaystyle\ \sum_{i=1}^{n}2i\ ## .

    And you have correctly stated that gives ##\displaystyle\ 2\left(\sum_{i=1}^{n}i\right)\ ##, so that is ##\displaystyle\ 2\left(\frac{n(n+1)}{2}\right) ## .

    What is the big sum, 1+2+3+4+5+(2n-1)+2n, i.e. ##\displaystyle\ \sum_{i=1}^{2n}i\ \ ? ##
  17. May 9, 2015 #16
    I can't figure this out, please help. I need a worked example I can refer to
  18. May 9, 2015 #17


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    If you understand the results shown above, I think we can get the correct result using a alternative method to Spivak's.

    Now consider the sum of the first n odd natural numbers, ##\displaystyle\ \sum_{i=1}^{n}(2i-1)\ ##.

    Split that up into ##\displaystyle\ \sum_{i=1}^{n}(2i)-\sum_{i=1}^{n}(1)\\ ##

    What do those last two sums give?
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