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Math Help - Steps between 2 forms of divisons

  1. #1
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    Steps between 2 forms of divisons

    Hello!

    (((l+1)^2)/((2l+1)*(2l+3))) + ((l*(l+1))/(2*(2l+1))) - Wolfram|Alpha

    I can't solve, how to get the Alternate form from the Input one on the link. Please somebody help me, step-by step to understand, how do the calculator did it.
    Thanks
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  2. #2
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    Re: Steps between 2 forms of divisons

    Quote Originally Posted by Creephun View Post
    Hello!

    (((l+1)^2)/((2l+1)*(2l+3))) + ((l*(l+1))/(2*(2l+1))) - Wolfram|Alpha

    I can't solve, how to get the Alternate form from the Input one on the link. Please somebody help me, step-by step to understand, how do the calculator did it.
    Thanks
    Here's a "road map" for you.

    I'm going to relabel your "l" since it looks so much like a "1". I'll call it x.
    It looks like
    \frac{(x + 1)^2}{(2x + 1)(2x + 3)} + \frac{x(x + 1)}{2(2x + 1)}

    I'm going to do a lot of factoring here. You can do it by the brute force method, but the factoring approach is a bit easier to see. So let's factor what we can here.

    \left [ \frac{x + 1}{2x + 1} \right ] \cdot \left [ \frac{x + 1}{2x + 3} + \frac{x}{2} \right ]

    Fractions and instructions to simplify. Let's get some common denominators and do another factorization.

    \left [ \frac{x + 1}{2(2x + 1)(2x + 3)} \right ] \cdot \left [ 2x + 2 + 2x^2 + 3x  \right ]

    Combining terms and factoring the numerator (on the right side brackets) gives
    \left [ \frac{x + 1}{2(2x + 1)(2x + 3)} \right ] \cdot \left [ (x + 2)(2x + 1)  \right ]

    and canceling the common terms we finally get
    \frac{(x + 1)(x + 2)}{2(2x + 3)}

    -Dan
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  3. #3
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    Re: Steps between 2 forms of divisons

    Thank you very much!
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