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Thread: Question Discrete Math Binomial Theorem

  1. #1
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    Question Question Discrete Math Binomial Theorem


    Please could someone help me the way to approach this?
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  2. #2
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    Re: Question Discrete Math Binomial Theorem

    The binomial theorem states (as you could have looked up)

    $(a+b)^n = \sum \limits_{k=0}^n ~a^k ~b^{n-k}$

    you should be able to figure out where to plug in $2$ and $(-1)$ into the formula above
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  3. #3
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    Re: Question Discrete Math Binomial Theorem

    Quote Originally Posted by romsek View Post
    The binomial theorem states (as you could have looked up)

    $(a+b)^n = \sum \limits_{k=0}^n ~a^k ~b^{n-k}$

    you should be able to figure out where to plug in $2$ and $(-1)$ into the formula above
    Not quite! You left out the binomial coefficient that was already in inayat's formula.

    The binomial theorem says that

    $(a+b)^n = \sum \limits_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} ~a^k ~b^{n-k}$

    Where $\begin{pmatrix}n \\ k\end{pmatrix}$ is the "binomial coefficient", $\begin{pmatrix}n \\ k\end{pmatrix}= \frac{n!}{k!(n-k)!}$.
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  4. #4
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    Re: Question Discrete Math Binomial Theorem

    Quote Originally Posted by inayat View Post

    Please could someone help me the way to approach this?
    I hope the is a clear proof for you.
    $\begin{gathered}
    \sum\limits_{k = 0}^n {\dbinom{n}{k}{a^{n - k}}{b^k} = {{(a + b)}^n}} \hfill \\
    \sum\limits_{k = 0}^n {\dbinom{n}{k}{2^{n - k}}{{( - 1)}^k} = {{(2 + ( - 1))}^n}} \hfill \\
    = 1 \hfill \\
    \end{gathered} $
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