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Math Help - Summing proof

  1. #1
    Super Member Deadstar's Avatar
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    Summing proof

    Let n be a nonnegative integer. Use the identity

    {(1 + x)^n}{(1 + x)^n} = {(1 + x)^{2n}}

    To show that

    \sum_{k=0}^n {n \choose k}^2 = {2n \choose n}.

    So far ive gotten to...

    Using the binomial theorem;

    (x + y)^n = \sum_{k=0}^n {n \choose k} x^{n-k}y^k and setting x and y to one we get;

    (1 + x)^n = \sum_{k=0}^n {n \choose k}

    so...

     \sum_{k=0}^n {n \choose k}^2 = (1 + x)^n(1 + x)^n = (1 + x)^{2n} but from here i dont know what to do... Any help please?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Deadstar View Post
    Let n be a nonnegative integer. Use the identity

    {(1 + x)^n}{(1 + x)^n} = {(1 + x)^{2n}}

    To show that

    \sum_{k=0}^n {n \choose k}^2 = {2n \choose n}.

    So far ive gotten to...

    Using the binomial theorem;

    (x + y)^n = \sum_{k=0}^n {n \choose k} x^{n-k}y^k and setting x and y to one we get;

    (1 + x)^n = \sum_{k=0}^n {n \choose k}

    so...

     \sum_{k=0}^n {n \choose k}^2 = (1 + x)^n(1 + x)^n = (1 + x)^{2n} but from here i dont know what to do... Any help please?
    I can't really help you, but you have two errors:
    (1 + x)^n = \sum_{k=0}^n {n \choose k}x^k

    And
     \sum_{k=0}^n \left ( {n \choose k} x^k \right )^2 \neq (1 + x)^n(1 + x)^n

    We can say that
     (1 + x)^n(1 + x)^n = \sum_{k = 0}^n \sum_{j = 0}^n {n \choose k}{n \choose j}x^kx^j
    but I don't see how this helps you.

    -Dan
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  3. #3
    Super Member Deadstar's Avatar
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    ok well how about

    \sum_{k=0}^n {n \choose k}^2 = (1 + 1)^n(1 + 1)^n = 2^{2n} where to now?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Deadstar View Post
    ok well how about

    \sum_{k=0}^n {n \choose k}^2 = (1 + 1)^n(1 + 1)^n = 2^{2n} where to now?
    This is what I'm trying to tell you
    \sum_{k=0}^n {n \choose k}^2 = (1 + 1)^n(1 + 1)^n \neq 2^{2n}

    Try it for n = 3 for example:
    \sum_{k=0}^n {n \choose k}^2 = 20
    not 2^{2 \cdot 3} = 2^6 = 64

    -Dan
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  5. #5
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    Using Danís hint:  (1 + x)^n(1 + x)^n = \sum_{k = 0}^n \sum_{j = 0}^n {n \choose k}{n \choose j}x^kx^j let x=1.
    Then  (2)^n(2)^n = \sum_{k = 0}^n \sum_{j = 0}^n {n \choose k}{n \choose j} and we know that  (2)^n = \sum_{k = 0}^n {n \choose k} .
    Can you finish?
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