sh3lsh
  • sh3lsh
How do I write this summation in its expanded form? Answer's given. Method needed.
Mathematics
chestercat
  • chestercat
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sh3lsh
  • sh3lsh
\[\sum_{j=2}^{n} \left(\begin{matrix}j \\ 2\end{matrix}\right) \to \frac{ n^{3} }{ 6 } - \frac{ n }{ 6 }\]
anonymous
  • 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.
sh3lsh
  • sh3lsh
Hey, you should get more credit for this. Thanks so much! It works out!

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anonymous
  • anonymous
Glad I could help :)
sh3lsh
  • sh3lsh
Real quick, how did you know to convert the combination to that?
anonymous
  • 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
anonymous
  • anonymous
Basically I just thought that I'd need to convert it into a more approachable form and tried it :)
sh3lsh
  • sh3lsh
Ah, duh! Thanks again!
anonymous
  • anonymous
No worries!

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