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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.

Thanks!

\[\sum_{k=1}^{n}(2+\frac{2 k }{ n })^2\]?

gotta square it first

so 4+2k^2/n?

oh no, square the whole thing

Could you help show me?

\[(2+\frac{2k}{n})(2+\frac{2k}{n})=4+\frac{8k}{n}+\frac{4k^2}{n^2}\]

then distribute the summation

now break it up in to three summations
\[\sum_{k=1}^n4=4n\] is the first one

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
\[\frac{8}{n}\sum_{k=1}^nk=\frac{8}{n}\frac{n(n+1)}{2}=4(n+1)\]

So is it (n(n+1)(n+2))/6

http://polysum.tripod.com/
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!! \(4(n+1)=4n+4\)!!

\[4n+4n+4+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]

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

copy and paste, you will see it , and see that it is right