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Math Help - Sum

  1. #1
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    Sum

    How to prove that \sum_{m=0}^n\binom{2n}{2m}9^{n-m}=\frac12\big(4^{2n}+2^{2n}\big) ?
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let S_1=\sum_{m=0}^nC_{2n}^{2m}9^{n-m}=\sum_{m=0}^nC_{2n}^{2n}3^{2n-2m}

    S_2=\sum_{m=1}^nC_{2n}^{2m-1}3^{2n-2m+1}

    Then

    S_1+S_2=(3+1)^{2n}=4^{2n}

    S_1-S_2=(3-1)^{2n}=2^{2n}

    Add the equalities:

    2S_1=4^{2n}+2^{2n}\Rightarrow S_1=\frac{1}{2}(4^{2n}+2^{2n}
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  3. #3
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    Thanks but, I don't see how did you get S_1+S_2=(3+1)^{2n} ? I can't follow that equality, I was trying to figure out by myself but I couldn't.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Soldado View Post
    Thanks but, I don't see how did you get S_1+S_2=(3+1)^{2n} ? I can't follow that equality, I was trying to figure out by myself but I couldn't.
    S_1+S_2 =\sum_{m=0}^{n} \left(\begin{array}{c}2n\\2m\end{array}\right) 3^{2n-2m} + \sum_{m=1}^n \left(\begin{array}{c}2n\\2m-1\end{array}\right) 3^{2n-2m+1}


    lats expand S1

    \sum_{m=0}^{n} \left(\begin{array}{c}2n\\2m\end{array}\right) 3^{2n-2m} = 3^{2n} + \left(\begin{array}{c}2n\\2\end{array}\right)3^{2n-2} + \left(\begin{array}{c}2n\\4\end{array}\right) 3^{2n-4} + ... +\left(\begin{array}{c}2n\\2n-2\end{array}\right)3^2

    S2 in the same way

    \sum_{m=1}^n \left(\begin{array}{c}2n\\2m-1\end{array}\right) 3^{2n-2m+1} = \left(\begin{array}{c}2n\\1\end{array}\right)3^{2n-1} + \left(\begin{array}{c}2n\\3\end{array}\right)3^{2n-3} +... + \left(\begin{array}{c}2n\\2n-1\end{array}\right)3

    S_1 + S_2 = 3^{2n} +\left(\begin{array}{c}2n\\1\end{array}\right)3^{2  n-1} +  \left(\begin{array}{c}2n\\2\end{array}\right)3^{2n-2} +  \left(\begin{array}{c}2n\\3\end{array}\right)3^{2n-3} +\left(\begin{array}{c}2n\\4\end{array}\right) 3^{2n-4}  +...+ \left(\begin{array}{c}2n\\2n-2\end{array}\right)3^2+ \left(\begin{array}{c}2n\\2n-1\end{array}\right)3

    this is the the expand of

    (1+3)^{2n}
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