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Math Help - Prove by using Binomial Theorem

  1. #1
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    Prove by using Binomial Theorem

    \sum i \binom{n}{i}=n2^{n-1}

    The summation is from i=0 to n but I couldnt figure out how to put that in with latex.

    I tried expanding out the Left Side but that didn't lead me anywhere.

    I thought about splitting the right side's 2 into a form of (x+y)^n but I couldn't figure out what my x and y should be.

    Help please?
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  2. #2
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    Quote Originally Posted by thelaughingman View Post
    \sum_{i=1}^n i \binom{n}{i}=n2^{n-1}
    Are you also allowed to use a bit of calculus? If so, write down the binomial expansion of (x+1)^n, differentiate it and then put x=1.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Are you also allowed to use a bit of calculus? If so, write down the binomial expansion of (x+1)^n, differentiate it and then put x=1.
    Damn it, I thought of splitting 2 into (1+1) but never thought of (x+1) and subbing x=1 later.

    I'm not sure if this is a valid solution but I'll find out tomorrow.

    Thanks a lot!
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  4. #4
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    Using the binomial theorem with a = b = 1, putting m = n-1 then multiplying both sides by n, we have:

    \displaystyle 2^m = \sum_{k=0}^{m}\binom{m}{k} \Rightarrow \displaystyle 2^{n-1} = \sum_{k=0}^{n-1}\binom{n-1}{k} \Rightarrow \displaystyle n2^{n-1} = \sum_{k=0}^{n-1}{n}\binom{n-1}{k} = \sum_{k=0}^{n}{k}\binom{n}{k}.
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