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Math Help - Proof on the summation of a finite series

  1. #1
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    Proof on the summation of a finite series

    Need help on the proof of \sum^{n}_{r=1}r^2=\frac{1}{6}n(n+1)(2n+1)

    The series of the squares of the first n natural numbers , ie

     \sum^{n}_{r=1}r^2=1^2+2^2+3^2+...+(n-1)^2+n^2

    I know this identity is needed to complete the proof :

     <br />
(r+1)^3=r^3+3r^3+3r+1 <br /> <br />
    From here onwards , i am not sure what to do ..


    Not sure how to proof this as well :

     <br />
\sum^{n}_{r=1}r^3=\frac{1}{4}n^2(n+1)^2<br />
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  2. #2
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    Hi

    There are several ways to prove that \sum^{n}_{r=1}r^2=\frac{1}{6}n(n+1)(2n+1)

    One is induction but obviously the beginning of your proof suggests another one.

    Write the equality from (r+1) down to 1
    (r+1)^3=r^3+3r^2+3r+1
    r^3=(r-1)^3+3(r-1)^2+3(r-1)+1
    etc ...
    2^3=1^3+3\cdot 1^2+3\cdot 1+1
    1^3=0^3+3\cdot 0^2+3\cdot 0+1

    Then sum all these equalities and simplify, some terms being on both sides
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  3. #3
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    Re :

    My book says :

    By using this identity

    (r+1)^3=r^3+3r^2+3r+1

    By adding the terms of the identity one by one with values of r from r=n to r=1 , we get

     \sum^{n}_{r=1}[(r+1)^3-r^3]=3\sum^{n}_{r=1}r^2+3\sum^{n}_{r=1}r+\sum^{n}_{r=1  }1

    The LHS side of this identity can be written as :

     [(n+1)^3-n^3]+[n^3-(n-1)^3]+...+(3^3-2^3)+(2^3-1^3)=(n+1)^3-1


    MY question is why is
     [(n+1)^3-n^3]+[n^3-(n-1)^3]+...+(3^3-2^3)+(2^3-1^3)=(n+1)^3-1
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by mathaddict View Post
    MY question is why is
     [(n+1)^3-n^3]+[n^3-(n-1)^3]+...+(3^3-2^3)+(2^3-1^3)=(n+1)^3-1
    Remove [ and ] and cancel positive and negative terms in LHS
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