A community for students.
Here's the question you clicked on:
 0 viewing
AJW99
 3 years ago
Pleas Help. Sigma notation
see equation below
AJW99
 3 years ago
Pleas Help. Sigma notation see equation below

This Question is Closed

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

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0Lemme double check what exactly closed form is and I may be able to assist.

uber1337h4xx0r
 3 years ago
Best ResponseYou've already chosen the best response.0Dang 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 :/

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1gotta square it first

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1oh no, square the whole thing

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

Mr.Me
 3 years ago
Best ResponseYou've already chosen the best response.0then distribute the summation

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

satellite73
 3 years ago
Best 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 \[\sum_{k=1}^nk\]?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1notice that \(n\) is fixed so it comes out of the summation

AJW99
 3 years ago
Best ResponseYou've already chosen the best response.0I am really lost on how to do sigmas. I really appreciate the help you are giving me

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

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

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1the 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1http://polysum.tripod.com/ yes it is!

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1so you get for the last term \[\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\] oops you were off by a little

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1should be \(2n+1\) in the numerator, not \(n+2\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1it is algebra from here on in

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1unless of course i messed up my algebra let me check

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i am off somewhere not sure exactly where

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1oh damn distributive law!! \(4(n+1)=4n+4\)!!

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1\[4n+4n+4+\frac{4}{n^2}\frac{n(n+1)(2n+1)}{6}\]

AJW99
 3 years ago
Best ResponseYou've already chosen the best response.0I see now . Thank you so much for your help I really appreciate it!

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1here we can check what it looks like in a nice form

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1http://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 \[\frac{2(2n+1)(7n+1)}{3n}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1and 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1copy and paste, you will see it , and see that it is right
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.