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Thread: Maximum Values for N-Choose-K Relations

  1. #1
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    Maximum Values for N-Choose-K Relations

    What is the general method for finding the maximum values for N-Choose-K formulations?

    In particular, my problem is asking for the k that makes (100-Choose-k)*2^k the maximum value.

    I'm certain that the binomial theorem is required to tackle this problem, but the 2^k is throwing me off,

    although I do know that, (100 choose 0)+(100 choose 1)+...+(100 choose 100) is 2^100.

    Any help for me?
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  2. #2
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    Quote Originally Posted by buckleybees View Post
    What is the general method for finding the maximum values for N-Choose-K formulations?

    In particular, my problem is asking for the k that makes (100-Choose-k)*2^k the maximum value.

    I'm certain that the binomial theorem is required to tackle this problem, but the 2^k is throwing me off,

    although I do know that, (100 choose 0)+(100 choose 1)+...+(100 choose 100) is 2^100.

    Any help for me?
    As k increases, a_k = \textstyle{100\choose k}2^k will increase up to a certain value of k, and then decrease. The ratio \frac{a_{k+1}}{a_k} is equal to \frac{2(100-k)}{k+1}. Solve the inequality \frac{2(100-k)}{k+1}\geqslant1 to get k\leqslant66\tfrac13. Thus a_k increases until k=66, and then decreases. So k=66 gives the maximum.
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