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  1. #1
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    binomials...

    oh...my questions are challenge enough..wish someone can answer me this binomial theory...i really sucks at this..

    Show that,

    <br />
\sum_{k=0}^{n}\binom{k+n}{k}{2}^{-k}=2^n<br />
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  2. #2
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    Did you copy that correctly?

    If it's supposed to be \sum_{k=0}^{n+k}\binom{k+n}{k}{2}^{-k}=2^n


    This was already shown \sum_{k=0}^{n}\binom{n}{k}=2^n

    So, \sum_{k=0}^{n+k}\binom{n+k}{k}=2^{n+k}

    This gives us \sum_{k=0}^{n+k}\binom{n+k}{k}2^{-k}=2^{n}
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  3. #3
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    Quote Originally Posted by Bingk View Post
    Did you copy that correctly?

    If it's supposed to be {\color{red}\sum_{k=0}^{n+k}}\binom{k+n}{k}{2}^{-k}=2^n

    This was already shown \sum_{k=0}^{n}\binom{n}{k}=2^n

    So, \sum_{k=0}^{n+k}\binom{n+k}{k}=2^{n+k}

    This gives us \sum_{k=0}^{n+k}\binom{n+k}{k}2^{-k}=2^{n}
    Sorry to have to say so, but the expression in red makes no sense (nor does the rest of that argument). if k is the summation index, then it cannot also occur as part of one of the limits of summation. The summation index is a variable, and the limits of summation must be constants.

    Quote Originally Posted by manusform View Post
    Show that

    <br />
\sum_{k=0}^{n}\binom{k+n}{k}{2}^{-k}=2^n<br />
    The left side of that identity is the coefficient of x^n in the expression \sum_{k=0}^{n}\bigl(x+\tfrac12\bigr)^{n+k}. But that is a geometric series, with sum \frac{\bigl(x+\tfrac12\bigr)^{2n+1} - \bigl(x+\tfrac12\bigr)^n}{x-\tfrac12}. To find the coefficient of x^n there, write \frac1{x-\tfrac12} = -2(1-2x)^{-1} = -2\bigl(1+2x+(2x)^2+(2x)^3+\ldots\bigr).

    Picking out the coefficient of x^n in -2\bigl(\bigl(x+\tfrac12\bigr)^{2n+1} - \bigl(x+\tfrac12\bigr)^n\bigr) \bigl(1+2x+(2x)^2+(2x)^3+\ldots\bigr), you find that it is

    \begin{aligned} -2\biggl(&{2n+1\choose n}2^{-n-1} + {2n+1\choose n-1}2^{-n-1} + {2n+1\choose n}2^{-n-1} + \ldots + {2n+1\choose 0}2^{-n-1} \\ &  - 1 - {n\choose 1}- \ldots - {n\choose n}\biggr).\end{aligned}

    But in the second line of that expression, the sum of all the binomial coefficients \textstyle {n\choose k} is 2^n. In the first line, the sum of half of the binomial coefficients \textstyle {2n+1\choose k} (with k going from 0 to n) is \tfrac12\cdot2^{2n+1} = 2^{2n}. That gives the value for the coefficient of x^n as -2(2^{n-1} - 2^n) = 2^n.
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  4. #4
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    Yup, you're absolutely right, I completely missed that ... sorry.
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