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sh3lsh

  • one year ago

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

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  1. sh3lsh
    • one year ago
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    \[\sum_{j=2}^{n} \left(\begin{matrix}j \\ 2\end{matrix}\right) \to \frac{ n^{3} }{ 6 } - \frac{ n }{ 6 }\]

  2. anonymous
    • one year ago
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    \[\sum_{j=2}^{n}\left(\begin{matrix}j \\ 2\end{matrix}\right) = \sum_{j=2}^{n}\frac{ j(j-1) }{ 2 } = \frac{ 1 }{ 2 }\sum_{j=2}^{n}j^{2} - \frac{ 1 }{ 2 }\sum_{j=2}^{n}j\] 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
    • one year ago
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    Hey, you should get more credit for this. Thanks so much! It works out!

  4. anonymous
    • one year ago
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    Glad I could help :)

  5. sh3lsh
    • one year ago
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    Real quick, how did you know to convert the combination to that?

  6. anonymous
    • one year ago
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    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
    • one year ago
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    Basically I just thought that I'd need to convert it into a more approachable form and tried it :)

  8. sh3lsh
    • one year ago
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    Ah, duh! Thanks again!

  9. anonymous
    • one year ago
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    No worries!

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