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Thread: simplifying a statement

  1. #1
    guenchy
    Guest

    simplifying a statement

    $\displaystyle (a(a+1) (2a + 1) )/ (6 ) + (a+1)^2 = ([a+1]([a+1]+1) (2[a+1]+1) )/ (6)$



    also like this

    a(a+1) (2a + 1) ------------- ---[a+1]([a+1]+1) (2[a+1]+1)
    ----------------+ (a+1)^2 =--------------------------
    --------6 -------------------------------- 6
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  2. #2
    Super Member

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    Hello, guenchy!

    This seems to be an Inducation problem.

    Something like . . .
    . . Prove by induction that: .$\displaystyle 1^2 + 2^2 + 3^2 + \cdots n^2 \;=\;\frac{n(n+1)(2n+1)}{6}$


    $\displaystyle \frac{a(a+1) (2a + 1)}{6} + (a+1)^2 \;= \;\frac{(a+1)([a+1]+1) (2[a+1]+1)}{6} \;=$ . $\displaystyle {\color{blue}\frac{(a+1)(a+2)(2a+3)}{6}}$
    You've verified $\displaystyle S(1)$.

    You assumed $\displaystyle S(a)\!:\;1^2+ 2^2 + 3^2 + a^2 \;=\;\frac{a(a+1)(2a+1)}{6}$

    You added $\displaystyle (a+1)^2$ to both sides:
    . . $\displaystyle 1^2 + 2^2 + 3^2 + \cdots + a^2 + (a+1)^2 \;=\;\frac{a(a+1)(2a+1)}{6} + (a+1)^2$

    Now we want to show that the right side is the right side of $\displaystyle S(a+1)$,

    . . which is: .$\displaystyle \frac{(a+1)([a+1]+1)(2[a+1]+1)}{6} \;=\;\frac{(a+1)(a+2)(2a+3)}{6}$


    Simplify the left side . . .
    It's just algebra, isn't it?

    . . $\displaystyle \frac{a(a+1)(2a+1)}{6} + (a+1)^2 \;=\;\frac{a(a+1)(2a+1) + 6(a+1)^2}{6}$

    Factor: .$\displaystyle \frac{(a+1)\,[a(2a+1) + 6(a+1)]}{6} \;=\;\frac{(a+1)(2a^2 + a + 6a + 6)}{6}$

    . . $\displaystyle = \;\frac{(a+1)(2a^2+7a+6)}{6} \;=\;\frac{(a+1)(a+2)(2a+3)}{6}$ . . . . There!

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