anonymous
  • anonymous
Pleas Help. Sigma notation see equation below
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Express the following in closed form \[\sum_{k=1}^{n}(2+2*\frac{ k }{ n })^2\]
anonymous
  • anonymous
Lemme double check what exactly closed form is and I may be able to assist.
anonymous
  • anonymous
Thanks!

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anonymous
  • anonymous
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 :/
anonymous
  • anonymous
\[\sum_{k=1}^{n}(2+\frac{2 k }{ n })^2\]?
anonymous
  • anonymous
gotta square it first
anonymous
  • anonymous
so 4+2k^2/n?
anonymous
  • anonymous
oh no, square the whole thing
anonymous
  • anonymous
Could you help show me?
anonymous
  • anonymous
\[(2+\frac{2k}{n})(2+\frac{2k}{n})=4+\frac{8k}{n}+\frac{4k^2}{n^2}\]
anonymous
  • anonymous
then distribute the summation
anonymous
  • anonymous
now break it up in to three summations \[\sum_{k=1}^n4=4n\] is the first one
anonymous
  • anonymous
\[\sum_{k=1}^n\frac{8k}{n}=\frac{8}{n}\sum_{k=1}^nk\] for the second do you know how to add \[\sum_{k=1}^nk\]?
anonymous
  • anonymous
notice that \(n\) is fixed so it comes out of the summation
anonymous
  • anonymous
I am really lost on how to do sigmas. I really appreciate the help you are giving me
anonymous
  • anonymous
\[\frac{4 n(n+1)(2n+1) }{ 6n ^{2} }\]
anonymous
  • anonymous
the formula for \[\sum_{k=1}^nk=1+2+3+...+n=\frac{n(n+1)}{2}\]
anonymous
  • anonymous
so the second term gives \[\frac{8}{n}\sum_{k=1}^nk=\frac{8}{n}\frac{n(n+1)}{2}=4(n+1)\]
anonymous
  • anonymous
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
anonymous
  • anonymous
So is it (n(n+1)(n+2))/6
anonymous
  • anonymous
http://polysum.tripod.com/ yes it is!
anonymous
  • anonymous
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
anonymous
  • anonymous
should be \(2n+1\) in the numerator, not \(n+2\)
anonymous
  • anonymous
it is algebra from here on in
anonymous
  • anonymous
but one answer is \[4n+4n+1+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]
anonymous
  • anonymous
unless of course i messed up my algebra let me check
anonymous
  • anonymous
i am off somewhere not sure exactly where
anonymous
  • anonymous
I can't find it either
anonymous
  • anonymous
oh damn distributive law!! \(4(n+1)=4n+4\)!!
anonymous
  • anonymous
\[4n+4n+4+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]
anonymous
  • anonymous
I see now . Thank you so much for your help I really appreciate it!
anonymous
  • anonymous
here we can check what it looks like in a nice form
anonymous
  • anonymous
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 \[\frac{2(2n+1)(7n+1)}{3n}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
copy and paste, you will see it , and see that it is right

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