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    Hello I know that the summation of r^2 from r=1 to r=n is [n(n+1)(2n+1)]/6, but I wasn't able to derive it. My approach was to write it as 1,1+2+1,1+4+4,1+6+9... but it didn't get me anywhere. A bit of aid would be much appreciated.
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    (Original post by Spandy)
    Hello I know that the summation of r^2 from r=1 to r=n is [n(n+1)(2n+1)]/6, but I wasn't able to derive it. My approach was to write it as 1,1+2+1,1+4+4,1+6+9... but it didn't get me anywhere. A bit of aid would be much appreciated.
    there are several different proofs

    e.g induction

    or

    method of differences with a choice of suitable identities
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    Oh crap, induction had totally slipped out of my mind, thanks!!
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    (Original post by Spandy)
    Oh crap, induction had totally slipped out of my mind, thanks!!
    my pleasure
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    Do you know how to derive Σk? Whatever method you used for this also applies here!
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    (Original post by NoNewFriends)
    Do you know how to derive Σk? Whatever method you used for this also applies here!
    No it doesn't

    Unless I have completely misunderstood what you are suggesting
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    (Original post by TenOfThem)
    No it doesn't

    Unless I have completely misunderstood what you are suggesting
    You know a lot more than I know so I'm probably wrong but which method do you have in mind that works for Σk but doesn't for Σk²?
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    (Original post by NoNewFriends)
    You know a lot more than I know so I'm probably wrong but which method do you have in mind that works for Σk but doesn't for Σk²?
    Standard proof for the sum of the first n numbers
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    (Original post by TenOfThem)
    Standard proof for the sum of the first n numbers
    Which one is the standard one?
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    (Original post by NoNewFriends)
    Which one is the standard one?
    Write the numbers in reverse
    Add to the original series
    Half
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    (Original post by NoNewFriends)
    Which one is the standard one?
    Let:
    S_n= \displaystyle\sum_{k=1}^n k
    Then
    S_n=1+2+. . .+(n-1)+n

\Rightarrow S_n=n+(n-1)+. . .+2 +1

\Rightarrow 2S_n=\underbrace{(n+1)+(n+1)+. . .+(n+1)+(n+1)}_{\text{n \ times}}

\Rightarrow S_n=\dfrac{n(n+1)}{2}
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    (Original post by NoNewFriends)
    Which one is the standard one?
    Hurrah for Joostan. Or simply use the formula for arithmetic progressions
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    I wrote this a long time back on my website (http://www.whitegroupmaths.com/2010/...hematical.html see Q6), hopefully it helps. Peace.

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    I wrote this a long time back on my website (http://www.whitegroupmaths.com/2010/...hematical.html see Q6), hopefully it helps. Peace.

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    (Original post by Spandy)
    Hurrah for Joostan. Or simply use the formula for arithmetic progressions
    You were looking for proof
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    (Original post by TenOfThem)
    You were looking for proof
    Uhh, I must be drunk.
    My bad...
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    (Original post by WhiteGroupMaths)
    I wrote this a long time back on my website (http://www.whitegroupmaths.com/2010/...hematical.html see Q6), hopefully it helps. Peace.
    Or similarly but with r(r+1)(r+2) - (r-1)r(r+1) = 3r(r+1).
 
 
 
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