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

Question

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

sh3lsh · Answer

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

anonymous · Answer

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

sh3lsh · Answer

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

anonymous · Answer

Glad I could help :)

sh3lsh · Answer

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

anonymous · Answer

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

anonymous · Answer

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

sh3lsh · Answer

Ah, duh! 
Thanks again!

anonymous · Answer

No worries!