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Mendicant_Bias

  • one year ago

(Introductory Real Analysis) - This isn't actually the problem I'm being asked to do, but it's part of it; I'm trying to find the formula for the sum of all even integers, and the way I'm doing it conceptually makes sense to me, but isn't working out mathematically. Why?

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  1. Mendicant_Bias
    • one year ago
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    What I'm doing is taking the formulas for both the summation of all positive integers and the summation of all even positive integers and subtracting one from the other, to hopefully yield the summation formula for the sum of all odd positive integers. Why isn't this working? \[S_{N(even)}=n(n+1)\]\[S_{N(all)}=\frac{n(n+1)}{2}\]\[S_{N(all)}-S_{N(even)}=\frac{n(n+1)}{2}-n(n+1)=\frac{-n(n+1)}{2}=-\frac{n^2}{2}-\frac{n}{2}\] The formula for the sum of all odd digits is n^2, what am I doing wrong?

  2. ganeshie8
    • one year ago
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    How do you know the sum of even numbers is \(n(n+1)\) ?

  3. ganeshie8
    • one year ago
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    don't you think it must depend on whether \(n\) is odd or even ?

  4. Mendicant_Bias
    • one year ago
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    It looks like my formula is wrong then, yeah. Don't know how I missed that.

  5. anonymous
    • one year ago
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    S_N even is infinite set ?

  6. anonymous
    • one year ago
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    for finite set of even that start from 1 \(S_N=\{2,4,6,8,.....,2n\}=2\{1,2,3,....,n\} \) \(\sum S_N=2(n)(n+1)/2\) so if this is what you mean in ur formula then it's correct

  7. anonymous
    • one year ago
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    now what you have doing is wrong that is sum of all integers between 2n and 1 is \(\sum S_A=(2n)(2n+1)/2\)

  8. anonymous
    • one year ago
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    hmm it depend on how the last element looks like see in the even case last number only can be even.

  9. anonymous
    • one year ago
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    sum of odd \(S_O\) \(\sum S_O =2n^2+n-(n^2-n)=n^2\)

  10. anonymous
    • one year ago
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    :O i never knew sum of odds is n^2 weird !!!

  11. anonymous
    • one year ago
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    so what you have done wrong is note sum of all integers should be bigger than sum of only evens but look at ur summation \(S_{N(all)}=\frac{n(n+1)}{2}<n(n+1)=S_{N(even)}\) which is a contradiction and don't make sense :D

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