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Math Help - A Product Series Equation

  1. #1
    Newbie Feryll's Avatar
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    Question A Product Series Equation

    Before I get started, I am not sure if this should be in pre-calculus or here, so please forgive me if it is in the wrong place.
    Anyhow, I came upon this supposed equality in Picture Num 1.
    I originally found it in Picture Number 2.
    Could someone please explain why the equality in Picture Number One exists? Like what law states that it is true? Thanks
    P.S. If you can't see, the expression on the right of the sigma in Picture Number 2 is L^4
    Attached Thumbnails Attached Thumbnails A Product Series Equation-odd-equation.jpg   A Product Series Equation-picture-number-2.jpg  
    Last edited by Feryll; November 26th 2009 at 06:25 AM. Reason: Oh yes, I forgot I was asking a question
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  2. #2
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    Grandad's Avatar
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    Hello Feryll

    Welcome to Math Help Forum!
    Quote Originally Posted by Feryll View Post
    Before I get started, I am not sure if this should be in pre-calculus or here, so please forgive me if it is in the wrong place.
    Anyhow, I came upon this supposed equality in Picture Num 1.
    I originally found it in Picture Number 2.
    Could someone please explain why the equality in Picture Number One exists? Like what law states that it is true? Thanks
    P.S. If you can't see, the expression on the right of the sigma in Picture Number 2 is L^4
    Are you familiar with Proof by Induction? This is quite a straightforward appliction of this method.

    Suppose that P(x) is the propositional function \prod_{i=3}^x\left(1+\frac{3}{i-3}\right)=\frac{x^3-3x^2+2x}{6}=\frac{x(x-1)(x-2)}{6}

    Then P(x) \Rightarrow \prod_{i=3}^{x+1}\left(1+\frac{3}{i-3}\right)=\frac{x(x-1)(x-2)}{6}\cdot\left(1+\frac{3}{x-2}\right)
    =\frac{x(x-1)(x-2)}{6}\cdot\left(\frac{x+1}{x-2}\right)

    =\frac{[x+1]([x+1]-1)([x+1]-2)}{6}

    \Rightarrow P(x+1)
    Now P(4) is \prod_{i=4}^4\left(1+\frac{3}{i-3}\right)=\frac{(4)(3)(2)}{6}, which is true.

    So P(x) is true for all x \ge 4.

    Grandad
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  3. #3
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    Quote Originally Posted by Feryll View Post
    Before I get started, I am not sure if this should be in pre-calculus or here, so please forgive me if it is in the wrong place.
    Anyhow, I came upon this supposed equality in Picture Num 1.
    I originally found it in Picture Number 2.
    Could someone please explain why the equality in Picture Number One exists? Like what law states that it is true? Thanks
    P.S. If you can't see, the expression on the right of the sigma in Picture Number 2 is L^4
    1+ \frac{3}{i-3} = \frac{i}{i-3},
    so
    \prod_{i=4}^x \left(1 + \frac{3}{i-3} \right) = \frac{4}{1} \cdot \frac{5}{2} \cdot \frac{6}{3} \cdot \frac{7}{4} \cdots \frac{x-3}{x-6} \cdot \frac{x-2}{x-5} \cdot \frac{x-1}{x-4} \cdot \frac{x}{x-3}

    In the numerators of the fractions on the RHS, all but the last three, x-2, x-1, and x, cancel with the denominators of the preceding fractions. In the numerators, all but the first three, 1, 2, and 3, cancel with the denominators. So

    \prod_{i=4}^x \left(1 + \frac{3}{i-3} \right) = \frac{(x-2)(x-1)(x)}{1\cdot 2 \cdot 3} = (1/6) \; (x^3 - 3x^2 + 2x)

    [Edit]Beaten to the punch by Grandad! But my solution is different anyway.[/Edit]
    Last edited by awkward; November 26th 2009 at 07:49 AM. Reason: sour grapes
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