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How they get to this formula

  1. Mar 13, 2009 #1
    [tex]
    1^2+2^2+..+k^2=\frac{k(k+1)(k+2)}{6}
    [/tex]

    how to get this general result
    ??

    i know

    1+2+..+n=\frac{n(n+1)}{2}

    so by that rule the sum should be
    [tex]
    \frac{n(n^2+1)}{2}
    [/tex]
     
  2. jcsd
  3. Mar 13, 2009 #2

    statdad

    User Avatar
    Homework Helper

    Two methods come quickly to mind.

    First, since you are given that the sum should equal

    [tex]
    \frac{k (k+1)(k+2)}{6}
    [/tex]

    prove the result by induction.

    Alternatively, in general the sum

    [tex]
    1 + 2^r + 3^r + \dots + k^r
    [/tex]

    for integer [tex] r [/tex] is a polynomial of degree [tex] r + 1 [/tex]. For [tex] r = 2[/tex]

    [tex]
    1 + 2^2 + 3^2 + \dots + k^2 = ak^3 + bk^2 + ck + d
    [/tex]

    Evaluate the left side for four different values of [tex] k [/tex] (0, 1, 2, 3) and solve the system of linear equations for the coefficients.
     
  4. Mar 13, 2009 #3
    Or you can use the teleschoping series, to prove that.

    i.e

    [tex]\sum_{i=1}^{k}i^2=\frac{k(k+1)(k+2)}{6}[/tex]
     
  5. Mar 13, 2009 #4
    i cant understand how you get the equations out of you second alternative methos
     
  6. Mar 13, 2009 #5

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    You can't, because it's not true. Try it with k=2. 12+22=5, 2*3*4/6=4.


    What makes you think that? Hint: It's not valid either.
     
  7. Mar 13, 2009 #6

    Mentallic

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    Homework Helper

    Thanks, I found this very interesting :smile:


    Through the method shown by statdad, this sum is actually:

    (pending for you to find out)
     
    Last edited: Mar 13, 2009
  8. Mar 13, 2009 #7

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Bingo! :smile: But you shouldn't give answers away. :frown:
     
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