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Math Help - simplifying a statement

  1. #1
    guenchy
    Guest

    simplifying a statement

    (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: . 1^2 + 2^2 + 3^2 + \cdots n^2 \;=\;\frac{n(n+1)(2n+1)}{6}


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

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

    You added (a+1)^2 to both sides:
    . . 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 S(a+1),

    . . which is: . \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?

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

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

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

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