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Formula for 1^2 + 2 ^2 + +n^2?

  1. Sep 14, 2010 #1
    Formula for 1^2 + 2 ^2 +.... +n^2?

    Hello,

    I know how to figure out the formula for 1 +2 + 3 + ... + n (i.e., (1+n)n/2 ). But what is the formula for 1^2 + 2 ^2 + 3^2 + .... + n^2? How does one formulate it?

    Thanks,
    DDTHAI
     
  2. jcsd
  3. Sep 14, 2010 #2

    CRGreathouse

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    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    It's a cubic equation with rational coeffcients. You can find these by fitting the curve to the first four points.
     
  4. Sep 14, 2010 #3
    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    1^2 + 2^2 + 3^2 + ... + 2^n = 2^(n-1)
     
  5. Sep 14, 2010 #4
    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    sorry i don't read well the question.
     
  6. Sep 14, 2010 #5
    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    we know than (n-1)^3 = n^3 - 3n^2 + 3n -1. so

    0^3 = 1^3 - 3.1^2 + 3.1 - 1
    1^3 = 2^3 - 3.2^2 + 3.2 - 1
    2^3 = 3^3 - 3.3^2 + 3.3 - 1
    .
    .
    .
    (n-2)^3 = (n-1)^3 - 3(n-1)^2 + 3(n-1) -1
    (n-1)^3 = n^3 - 3n^2 + 3n - 1

    Then, if we plus all, we get
    n n n
    0^3 = n^3 - 3E (i)^2 + 3E n - E 1
    i=1 i=1 i=1

    n n n
    =>3E (i)^2 = n^3 + 3E n - E 1
    i=1 i=1 i=1


    n
    => 3E (i)^2 = n^3 + 3(n+1)n - n
    i=1 2

    After simplify you get

    n
    E (i)^2 = n(n+1)(2n+1)
    i=1 6

    *you must know than
    n
    E (i) = 1 + 2 `+ 3 + 4 + ... + n = n(n+1)
    i=1 2
     
  7. Sep 14, 2010 #6

    Mentallic

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    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    This makes absolutely no sense. I mean, look at it for a second... firstly you failed to notice the pattern correctly since 2^n means 2^1+2^2+2^3 instead of what is shown. And secondly, how can that all equal 2^(n-1) when on the left side of the equation, you already have 2^n which means the term just before the last is 2^(n-1)?

    Even though it was probably a typo or something, I just wanted to be annoying :tongue:
     
  8. Sep 14, 2010 #7
    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    See the book "ascent to orbit", A.C.Clarke's autobiography, chap 24. He list sums of powers up to 8.

    Sum n^2 = ( n / 6 )( n + 1 )( 2n + 2 )

    Sum n^3 = ( n^2 / 4 )( n + 1 )( n + 1 ) = [ sum n ]^2

    etc...
     
  9. Sep 14, 2010 #8
    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    -----------------------------------------------
    with n = 5, the series 1^2 + 2^2 + 3^2 + .... + n^2 = 1+4+9+16+25 = 55

    Sum n^2 = ( n / 6 )( n + 1 )( 2n + 2 ) = (5 / 6) * 6 * (2*5 + 2) = 5 * 12 = 60!
     
    Last edited: Sep 14, 2010
  10. Sep 14, 2010 #9
    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    Thanks Helios,

    I believe you tried to write
    Sum n^2 = ( n / 6 )( n + 1 )( 2n + 1 )

    That would be right for all n's that I have tried so far! But how do prove that the formula is right?

    Thanks you all for trying to help
    DDTHAI
     
  11. Sep 14, 2010 #10
    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    Thanks Little Ant - I now understand your approach for coming up with the formula!

    Once again, thank you all for helping.

    DDTHAI
     
  12. Sep 14, 2010 #11

    Gokul43201

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    Gold Member

    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    That is the correct formula. It can be proved by induction.
     
  13. Sep 14, 2010 #12
    Re: Formula for 1^2 + 2 ^2 +.... +n^2?

    Here is one method to derive the formula for the sum of squares which can be extended to other integer powers.

    Observe that [tex](k+1)^3 - k^3 = 3k^2 + 3k + 1[/tex] ... (*)

    Now sum up (*) for [tex]k = 1, 2, \dots , n[/tex].

    The sum on the left hand side telescopes, leaving [tex](n+1)^3 -1[/tex].

    The sum on the right hand side is
    [tex]3 \sum_{k=1}^n k^2 + 3 \sum_{k=1}^n k + n[/tex].

    Now equate these two expressions for the sum, apply the formula you already know for
    [tex]\sum_{k=1}^n k[/tex]
    and solve for
    [tex]\sum_{k=1}^n k^2[/tex].
     
    Last edited: Sep 14, 2010
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