## AJW99 Group Title Pleas Help. Sigma notation see equation below one year ago one year ago

1. AJW99 Group Title

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

2. uber1337h4xx0r Group Title

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

3. AJW99 Group Title

Thanks!

4. uber1337h4xx0r Group Title

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 :/

5. satellite73 Group Title

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

6. satellite73 Group Title

gotta square it first

7. AJW99 Group Title

so 4+2k^2/n?

8. satellite73 Group Title

oh no, square the whole thing

9. AJW99 Group Title

Could you help show me?

10. satellite73 Group Title

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

11. Mr.Me Group Title

then distribute the summation

12. satellite73 Group Title

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

13. satellite73 Group Title

$\sum_{k=1}^n\frac{8k}{n}=\frac{8}{n}\sum_{k=1}^nk$ for the second do you know how to add $\sum_{k=1}^nk$?

14. satellite73 Group Title

notice that $$n$$ is fixed so it comes out of the summation

15. AJW99 Group Title

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

16. Mr.Me Group Title

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

17. satellite73 Group Title

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

18. satellite73 Group Title

so the second term gives $\frac{8}{n}\sum_{k=1}^nk=\frac{8}{n}\frac{n(n+1)}{2}=4(n+1)$

19. satellite73 Group Title

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

20. AJW99 Group Title

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

21. satellite73 Group Title

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

22. satellite73 Group Title

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

23. satellite73 Group Title

should be $$2n+1$$ in the numerator, not $$n+2$$

24. satellite73 Group Title

it is algebra from here on in

25. satellite73 Group Title

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

26. satellite73 Group Title

unless of course i messed up my algebra let me check

27. satellite73 Group Title

i am off somewhere not sure exactly where

28. AJW99 Group Title

I can't find it either

29. satellite73 Group Title

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

30. satellite73 Group Title

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

31. AJW99 Group Title

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

32. satellite73 Group Title

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

33. satellite73 Group Title

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 $\frac{2(2n+1)(7n+1)}{3n}$

34. satellite73 Group Title

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

35. satellite73 Group Title

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