Mendicant_Bias
  • Mendicant_Bias
(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?
Mathematics
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schrodinger
  • schrodinger
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Mendicant_Bias
  • Mendicant_Bias
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?
ganeshie8
  • ganeshie8
How do you know the sum of even numbers is \(n(n+1)\) ?
ganeshie8
  • ganeshie8
don't you think it must depend on whether \(n\) is odd or even ?

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Mendicant_Bias
  • Mendicant_Bias
It looks like my formula is wrong then, yeah. Don't know how I missed that.
anonymous
  • anonymous
S_N even is infinite set ?
anonymous
  • anonymous
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
anonymous
  • anonymous
now what you have doing is wrong that is sum of all integers between 2n and 1 is \(\sum S_A=(2n)(2n+1)/2\)
anonymous
  • anonymous
hmm it depend on how the last element looks like see in the even case last number only can be even.
anonymous
  • anonymous
sum of odd \(S_O\) \(\sum S_O =2n^2+n-(n^2-n)=n^2\)
anonymous
  • anonymous
:O i never knew sum of odds is n^2 weird !!!
anonymous
  • anonymous
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}

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