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  1. #1
    Member SengNee's Avatar
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    Improve

    Question:
    If $\displaystyle f(r)=r(r+1)(r+2) $, find $\displaystyle f(r+1)-f(r)$ and deduce that $\displaystyle \sum_{r=1}^nr^2=\frac{1}{6}n(n+1)(2n+1)$

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

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

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



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






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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
    $\displaystyle \sum_{r=1}^nu_r=(n+1)( n+2)( n+3)-2(3)$

    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
    $\displaystyle \sum_{r=1}^nu_r=(n+1)( n+2)( n+3)-2(3)$

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

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

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