A community for students.
Here's the question you clicked on:
 0 viewing
sh3lsh
 one year ago
How do I write this summation in its expanded form?
Answer's given. Method needed.
sh3lsh
 one year ago
How do I write this summation in its expanded form? Answer's given. Method needed.

This Question is Closed

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{j=2}^{n} \left(\begin{matrix}j \\ 2\end{matrix}\right) \to \frac{ n^{3} }{ 6 }  \frac{ n }{ 6 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{j=2}^{n}\left(\begin{matrix}j \\ 2\end{matrix}\right) = \sum_{j=2}^{n}\frac{ j(j1) }{ 2 } = \frac{ 1 }{ 2 }\sum_{j=2}^{n}j^{2}  \frac{ 1 }{ 2 }\sum_{j=2}^{n}j\] From there you can use the identities for the sum of squares and integers and hopefully that gives you the right answer. If you don't know the identities there are ways to derive them.

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0Hey, you should get more credit for this. Thanks so much! It works out!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Glad I could help :)

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0Real quick, how did you know to convert the combination to that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well the formula for combinations is \[\left(\begin{matrix}n \\ k\end{matrix}\right)=\frac{ n! }{ k!(nk)! }\] I just immediately thought of that and applied it, turns out it worked :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Basically I just thought that I'd need to convert it into a more approachable form and tried it :)
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.