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    Where does that extra bit come from?
    I understand that the sequence starts from the 0th term so obviously standard results which start from the 1st term won't be just what you need. You need an extra little bit which is that underlined bit but how do you get it?
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    (Original post by will'o'wisp)
    Where does that extra bit come from?
    I understand that the sequence starts from the 0th term so obviously standard results which start from the 1st term won't be just what you need. You need an extra little bit which is that underlined bit but how do you get it?
    Well the first two terms are straight forward. And for the one's you're unsure:

    \displaystyle \sum_{r=1}^n (n) =\underbrace{n+n+n+n+...}_{n \text{times}} = n(n)

    \displaystyle \Rightarrow \sum_{r=0}^n (n) = n(n)+n=n(n+1)

    \displaystyle \Rightarrow \sum_{r=0}^n (2n) = 2n(n+1)

    On the second line, since you are now starting from r=0 instead of r=1, you include an extra n term.

    For \displaystyle \sum_{r=0}^n (1)=n+1 it's the exact same thought process.
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    (Original post by RDKGames)
    Well the firs two terms are straight forward.

    \displaystyle \sum_{r=1}^n (n) =\underbrace{n+n+n+n+...}_{n \text{times}} = n(n)

    \displaystyle \Rightarrow \sum_{r=0}^n (n) = n(n)+n=n(n+1)

    \displaystyle \Rightarrow \sum_{r=0}^n (2n) = 2n(n+1)

    On the second line, since you are now starting from r=0 instead of r=1, you include an extra n term.

    For \displaystyle \sum_{r=0}^n (1)=n+1 it's the exact same thought process.
    thanks i understand you need the terms from r=1 and then just add on the 0th term
 
 
 
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