Pleas Help. Sigma notation
see equation below

- anonymous

Pleas Help. Sigma notation
see equation below

- schrodinger

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- anonymous

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

- anonymous

Lemme double check what exactly closed form is and I may be able to assist.

- anonymous

Thanks!

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## More answers

- 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

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

- anonymous

gotta square it first

- anonymous

so 4+2k^2/n?

- anonymous

oh no, square the whole thing

- anonymous

Could you help show me?

- anonymous

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

- anonymous

then distribute the summation

- anonymous

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

- 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

notice that \(n\) is fixed so it comes out of the summation

- anonymous

I am really lost on how to do sigmas. I really appreciate the help you are giving me

- anonymous

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

- anonymous

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

- 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

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

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

- anonymous

http://polysum.tripod.com/
yes it is!

- 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

should be \(2n+1\) in the numerator, not \(n+2\)

- anonymous

it is algebra from here on in

- anonymous

but one answer is
\[4n+4n+1+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]

- anonymous

unless of course i messed up my algebra
let me check

- anonymous

i am off somewhere not sure exactly where

- anonymous

I can't find it either

- anonymous

oh damn distributive law!! \(4(n+1)=4n+4\)!!

- anonymous

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

- anonymous

I see now . Thank you so much for your help I really appreciate it!

- anonymous

here we can check what it looks like in a nice form

- 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

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

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

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