sh3lsh one year ago How do I write this summation in its expanded form? Answer's given. Method needed.

1. sh3lsh

$\sum_{j=2}^{n} \left(\begin{matrix}j \\ 2\end{matrix}\right) \to \frac{ n^{3} }{ 6 } - \frac{ n }{ 6 }$

2. anonymous

$\sum_{j=2}^{n}\left(\begin{matrix}j \\ 2\end{matrix}\right) = \sum_{j=2}^{n}\frac{ j(j-1) }{ 2 } = \frac{ 1 }{ 2 }\sum_{j=2}^{n}j^{2} - \frac{ 1 }{ 2 }\sum_{j=2}^{n}j$ 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.

3. sh3lsh

Hey, you should get more credit for this. Thanks so much! It works out!

4. anonymous

5. sh3lsh

Real quick, how did you know to convert the combination to that?

6. anonymous

Well the formula for combinations is $\left(\begin{matrix}n \\ k\end{matrix}\right)=\frac{ n! }{ k!(n-k)! }$ I just immediately thought of that and applied it, turns out it worked :P

7. anonymous

Basically I just thought that I'd need to convert it into a more approachable form and tried it :)

8. sh3lsh

Ah, duh! Thanks again!

9. anonymous

No worries!