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Thread: 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 : $\displaystyle (x+y)^n=\sum_{k=0}^n\binom{n}{k} x^ky^{n-k}$

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

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

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

    For example : $\displaystyle \binom{n}{n-k}=\frac{n!}{(n-k)![n-(n-k)]!}=\ldots$
    Last edited by flyingsquirrel; Jun 29th 2008 at 08:30 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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    $\displaystyle \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c} n \\ k \\
    \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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