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Pleas Help. Sigma notation see equation below

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Express the following in closed form \[\sum_{k=1}^{n}(2+2*\frac{ k }{ n })^2\]
Lemme double check what exactly closed form is and I may be able to assist.

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Other answers:

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 :/
\[\sum_{k=1}^{n}(2+\frac{2 k }{ n })^2\]?
gotta square it first
so 4+2k^2/n?
oh no, square the whole thing
Could you help show me?
then distribute the summation
now break it up in to three summations \[\sum_{k=1}^n4=4n\] is the first one
\[\sum_{k=1}^n\frac{8k}{n}=\frac{8}{n}\sum_{k=1}^nk\] for the second do you know how to add \[\sum_{k=1}^nk\]?
notice that \(n\) is fixed so it comes out of the summation
I am really lost on how to do sigmas. I really appreciate the help you are giving me
\[\frac{4 n(n+1)(2n+1) }{ 6n ^{2} }\]
the formula for \[\sum_{k=1}^nk=1+2+3+...+n=\frac{n(n+1)}{2}\]
so the second term gives \[\frac{8}{n}\sum_{k=1}^nk=\frac{8}{n}\frac{n(n+1)}{2}=4(n+1)\]
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
So is it (n(n+1)(n+2))/6 yes it is!
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
should be \(2n+1\) in the numerator, not \(n+2\)
it is algebra from here on in
but one answer is \[4n+4n+1+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]
unless of course i messed up my algebra let me check
i am off somewhere not sure exactly where
I can't find it either
oh damn distributive law!! \(4(n+1)=4n+4\)!!
I see now . Thank you so much for your help I really appreciate it!
here we can check what it looks like in a nice form\frac{4}{n^2}\frac{n%28n%2B1%29%282n%2B1%29}{6} so one way to write this mess is \[\frac{2(2n+1)(7n+1)}{3n}\]
and we can also check that it is in fact correct\sum_{k%3D1}^{n}%282%2B\frac{2+k+}{+n+}%29^2
copy and paste, you will see it , and see that it is right

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