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  1. #1
    Member SengNee's Avatar
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    Question:
    If f(r)=r(r+1)(r+2) , find f(r+1)-f(r) and deduce that \sum_{r=1}^nr^2=\frac{1}{6}n(n+1)(2n+1)

    My solution:
    f(r+1)-f(r)
    =(r+1)(r+2)(r+3)-r(r+1)(r+2)
    =(r+1)(r+2)[(r+3)-r]
    =3(r+1)(r+2)
    =3r^2+9r+6

    Let,
    u_r= f(r+1)-f(r)
    u_r=3r^2+9r+6
    r^2=\frac{1}{3}u_r-3r-2

    \sum_{r=1}^nr^2
    =\sum_{r=1}^n(\frac{1}{3}u_r-3r-2)
    =\frac{1}{3}\sum_{r=1}^nu_r-3\sum_{r=1}^n r-\sum_{r=1}^n 2
    =\frac{1}{3}[(n+1)( n+2)( n+3)-2(3)]-3[\frac{n}{2}(n+1)]-2n
    =\frac{(n^2+3n+2)(n+3)}{3}-2-\frac{3n(n+1)}{2}-2n
    =\frac{(n^3+3n^2+3n^2+9n+2n+6)}{3}-2-\frac{3n^2+3n}{2}-2n
    =\frac{2n^3+12n^2+22n+12-12-9n^2-9n-12n}{6}
    =\frac{1}{6}(2n^3+3n^2+n)
    =\frac{1}{6}n(2n^2+3n+1)
    =\frac{1}{6}n(n+1)(2n+1) \quad Proven



    \color{blue}\text{Anyone have better or shorter solutions, please show me to improve my long steps.}






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  2. #2
    Super Member angel.white's Avatar
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    I don't have a better way, but I was following along for the purposes of learning, and I am confused how you got that
    \sum_{r=1}^nu_r=(n+1)( n+2)( n+3)-2(3)

    If you wouldn't mind explaining that step, I would be appreciative.
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  3. #3
    Member SengNee's Avatar
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    Quote Originally Posted by angel.white View Post
    I don't have a better way, but I was following along for the purposes of learning, and I am confused how you got that
    \sum_{r=1}^nu_r=(n+1)( n+2)( n+3)-2(3)

    If you wouldn't mind explaining that step, I would be appreciative.
    If, u_r=f(r+1)-f(r)

    Then, \sum_{r=1}^nu_r=f(n+1)-f(1)
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by SengNee View Post
    If, u_r=f(r+1)-f(r)

    Then, \sum_{r=1}^nu_r=f(n+1)-f(1)
    I see, I never learned that (or maybe I just forgot it)
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