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AJW99

  • 2 years ago

Pleas Help. Sigma notation see equation below

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  1. AJW99
    • 2 years ago
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    Express the following in closed form \[\sum_{k=1}^{n}(2+2*\frac{ k }{ n })^2\]

  2. uber1337h4xx0r
    • 2 years ago
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    Lemme double check what exactly closed form is and I may be able to assist.

  3. AJW99
    • 2 years ago
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    Thanks!

  4. uber1337h4xx0r
    • 2 years ago
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    Dang it... it appears as though closed form deals with diff eq (according to a quick google). I am only at Cal 2, so I think I'm going to have to sit this one out :/

  5. satellite73
    • 2 years ago
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    \[\sum_{k=1}^{n}(2+\frac{2 k }{ n })^2\]?

  6. satellite73
    • 2 years ago
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    gotta square it first

  7. AJW99
    • 2 years ago
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    so 4+2k^2/n?

  8. satellite73
    • 2 years ago
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    oh no, square the whole thing

  9. AJW99
    • 2 years ago
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    Could you help show me?

  10. satellite73
    • 2 years ago
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    \[(2+\frac{2k}{n})(2+\frac{2k}{n})=4+\frac{8k}{n}+\frac{4k^2}{n^2}\]

  11. Mr.Me
    • 2 years ago
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    then distribute the summation

  12. satellite73
    • 2 years ago
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    now break it up in to three summations \[\sum_{k=1}^n4=4n\] is the first one

  13. satellite73
    • 2 years ago
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    \[\sum_{k=1}^n\frac{8k}{n}=\frac{8}{n}\sum_{k=1}^nk\] for the second do you know how to add \[\sum_{k=1}^nk\]?

  14. satellite73
    • 2 years ago
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    notice that \(n\) is fixed so it comes out of the summation

  15. AJW99
    • 2 years ago
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    I am really lost on how to do sigmas. I really appreciate the help you are giving me

  16. Mr.Me
    • 2 years ago
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    \[\frac{4 n(n+1)(2n+1) }{ 6n ^{2} }\]

  17. satellite73
    • 2 years ago
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    the formula for \[\sum_{k=1}^nk=1+2+3+...+n=\frac{n(n+1)}{2}\]

  18. satellite73
    • 2 years ago
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    so the second term gives \[\frac{8}{n}\sum_{k=1}^nk=\frac{8}{n}\frac{n(n+1)}{2}=4(n+1)\]

  19. satellite73
    • 2 years ago
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    the last one is \[\sum_{k=1}^n\frac{4k^2}{n^2}\] and as in the previous case the \(\frac{4}{n^2}\) comes out front to give you \[\frac{8}{n^2}\sum_{k=1}^nk^2\] so what you need for this one is the formula for \[\sum_{k=1}^nk^2\] which i am going to guess you do not know, but it is probably in your book

  20. AJW99
    • 2 years ago
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    So is it (n(n+1)(n+2))/6

  21. satellite73
    • 2 years ago
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    http://polysum.tripod.com/ yes it is!

  22. satellite73
    • 2 years ago
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    so you get for the last term \[\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\] oops you were off by a little

  23. satellite73
    • 2 years ago
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    should be \(2n+1\) in the numerator, not \(n+2\)

  24. satellite73
    • 2 years ago
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    it is algebra from here on in

  25. satellite73
    • 2 years ago
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    but one answer is \[4n+4n+1+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]

  26. satellite73
    • 2 years ago
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    unless of course i messed up my algebra let me check

  27. satellite73
    • 2 years ago
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    i am off somewhere not sure exactly where

  28. AJW99
    • 2 years ago
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    I can't find it either

  29. satellite73
    • 2 years ago
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    oh damn distributive law!! \(4(n+1)=4n+4\)!!

  30. satellite73
    • 2 years ago
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    \[4n+4n+4+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]

  31. AJW99
    • 2 years ago
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    I see now . Thank you so much for your help I really appreciate it!

  32. satellite73
    • 2 years ago
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    here we can check what it looks like in a nice form

  33. satellite73
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=4n%2B4n%2B4%2B\frac{4}{n^2}\frac{n%28n%2B1%29%282n%2B1%29}{6} so one way to write this mess is \[\frac{2(2n+1)(7n+1)}{3n}\]

  34. satellite73
    • 2 years ago
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    and we can also check that it is in fact correct http://www.wolframalpha.com/input/?i=\sum_{k%3D1}^{n}%282%2B\frac{2+k+}{+n+}%29^2

  35. satellite73
    • 2 years ago
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    copy and paste, you will see it , and see that it is right

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