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AJW99 Group TitleBest ResponseYou've already chosen the best response.0
Express the following in closed form \[\sum_{k=1}^{n}(2+2*\frac{ k }{ n })^2\]
 one year ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.0
Lemme double check what exactly closed form is and I may be able to assist.
 one year ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.0
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 :/
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{k=1}^{n}(2+\frac{2 k }{ n })^2\]?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
gotta square it first
 one year ago

AJW99 Group TitleBest ResponseYou've already chosen the best response.0
so 4+2k^2/n?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
oh no, square the whole thing
 one year ago

AJW99 Group TitleBest ResponseYou've already chosen the best response.0
Could you help show me?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[(2+\frac{2k}{n})(2+\frac{2k}{n})=4+\frac{8k}{n}+\frac{4k^2}{n^2}\]
 one year ago

Mr.Me Group TitleBest ResponseYou've already chosen the best response.0
then distribute the summation
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
now break it up in to three summations \[\sum_{k=1}^n4=4n\] is the first one
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_{k=1}^n\frac{8k}{n}=\frac{8}{n}\sum_{k=1}^nk\] for the second do you know how to add \[\sum_{k=1}^nk\]?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
notice that \(n\) is fixed so it comes out of the summation
 one year ago

AJW99 Group TitleBest ResponseYou've already chosen the best response.0
I am really lost on how to do sigmas. I really appreciate the help you are giving me
 one year ago

Mr.Me Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{4 n(n+1)(2n+1) }{ 6n ^{2} }\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
the formula for \[\sum_{k=1}^nk=1+2+3+...+n=\frac{n(n+1)}{2}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
so the second term gives \[\frac{8}{n}\sum_{k=1}^nk=\frac{8}{n}\frac{n(n+1)}{2}=4(n+1)\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

AJW99 Group TitleBest ResponseYou've already chosen the best response.0
So is it (n(n+1)(n+2))/6
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
http://polysum.tripod.com/ yes it is!
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
should be \(2n+1\) in the numerator, not \(n+2\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
it is algebra from here on in
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
but one answer is \[4n+4n+1+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
unless of course i messed up my algebra let me check
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i am off somewhere not sure exactly where
 one year ago

AJW99 Group TitleBest ResponseYou've already chosen the best response.0
I can't find it either
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
oh damn distributive law!! \(4(n+1)=4n+4\)!!
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[4n+4n+4+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]
 one year ago

AJW99 Group TitleBest ResponseYou've already chosen the best response.0
I see now . Thank you so much for your help I really appreciate it!
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
here we can check what it looks like in a nice form
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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 \[\frac{2(2n+1)(7n+1)}{3n}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
copy and paste, you will see it , and see that it is right
 one year ago
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