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I Sum of the first n squares?

  1. Jan 18, 2017 #1
    I found a deduction to determinate de sum of the first n squares. However there is a part on it that i didn't understood.


    We use the next definition: [tex](k+1)^3 - k^3 = 3k^2 + 3k +1[/tex], then we define [tex]k= 1, ... , n[/tex] and then we sum...

    [tex]
    (n+1)^3 -1 = 3\sum_{k=0}^{n}k^{2} +3\sum_{k=0}^{n}k+ n
    [/tex]

    The left side of the equality is the one that i didn't understood. Why [tex](k+1)^3 - k^3[/tex] changes in that way?
     
  2. jcsd
  3. Jan 18, 2017 #2

    haruspex

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    ((n+1)3-n3) + (n3-(n-1)3) + ((n-1)3-(n-2)3) ....
     
  4. Jan 18, 2017 #3

    Orodruin

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    The first term for k is cancelled by the second term for k+1. This leaves the first term for k=n and the second for k=1.
     
  5. Jan 19, 2017 #4

    Stephen Tashi

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    A general definition is ## \triangle F(k) = F(k+1) - F(k) ##
    So ##\triangle ( k^3) = (k+1)^3 - k^3 ##

    A general trick is the "telescoping sum":
    ##\sum_{k=1}^n \triangle F(k) = (F(1+1) - F(1)) + ( F(2+1) - F(2)) + (F(3+1) - F(3)) + ...+ F(n+1) - F(n)##
    ## = ( F(2) - F(1)) + (F(3) - F(2)) + (F(4) - F(3)) + ... + (F(n+1) - F(n)) ##
    ## = -F(1) + (F(2) - F(2)) + (F(3) - F(3)) + ...+ (F(n) - F(n))+ F(n+1) ##
    ## = F(n+1) - F(1) ##

    So ## \sum_{k=1}^n \triangle k^3 = (n+1)^3 - 1^3##
     
  6. Jan 19, 2017 #5

    mathman

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    [itex](k+1)^3=k^3+3k^2+3k+1[/itex]. Expand the expression.
     
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