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Math Help - proof with binomials

  1. #1
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    proof with binomials

    Hi... my english is not the best (I speak spanish), but... I need help with some problems... Can you help me please? ^^... Is urgent!

    Demonstrate:

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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Mariposa View Post
    Hi... my english is not the best (I speak spanish), but...
    That's not a problem
    Demonstrate:

    For the first one recall the binomial theorem : (x+y)^n=\sum_{k=0}^n\binom{n}{k} x^ky^{n-k}

    As for any two integers n and k one has 1^k=1 and 1^{n-k}=1, one can write \sum_{k=0}^n\binom{n}{k} = \sum_{k=0}^n\binom{n}{k}1^k1^{n-k}=(1+1)^n (thanks to the binomial theorem) and get \sum_{k=0}^n\binom{n}{k}=2^n. It gives us \left[\sum_{k=0}^n\binom{n}{k}\right]^2=(2^n)^2=2^{2n}=(2^2)^n=4^n

    Using this idea, what is \sum_{k=0}^n\binom{n}{k}3^k ?

    For the last two questions use the definition of \binom{n}{k} : \binom{n}{k}=\frac{n!}{k!(n-k)!}

    For example : \binom{n}{n-k}=\frac{n!}{(n-k)![n-(n-k)]!}=\ldots
    Last edited by flyingsquirrel; June 29th 2008 at 09:30 AM.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    They were already solved in another forum.
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  4. #4
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    \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c}   n  \\   k  \\<br />
\end{array}} \right)} 3^k  = \left( {3 + 1} \right)^n  = 4^n  = \left( {2^n } \right)^2  = \left( {\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c}  n  \\   k  \\\end{array}} \right)} } \right)^2
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