Find the maximum value of a summation

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The maximum value of the summation defined as max_{0≤x≤k} ∑_{(l_1,l_2)∈A} (N!/(N-l_1-l_2)!l_1!l_2!) x^{l_1}(k-x)^{l_2}(1-k)^{N-l_1-l_2} can be derived by differentiating the function x^{a}(k-x)^{b} and setting the derivative to zero. The specific conditions for the pairs (l_1, l_2) are defined by the set A, where l_1 + 2l_2 = l. The derived expression for the maximum value is (k l_1/(l_1+l_2))^{l_1} (k l_2/(l_1+l_2))^{l_2}, although the challenge lies in the fact that each term of the summation is maximized at different values of x as l_1 and l_2 vary.

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sabbagh80
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Hi,
What is the maximum value of the given summation in terms of k, l and N ?
max_{0\leq x \leq k} \sum_{(l_1,l_2)\in A} \frac{N!}{(N-l_1-l_2)!l_1!l_2!} x^{l_1}(k-x)^{l_2}(1-k)^{N-l_1-l_2}
where A=\{(l_1,l_2)|l_1,l_2 \in \{0,1,2,...,N\} and l_1+2l_2=l\} and 0<k<1.
Thanks a lot for your participation.
 
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Forgetting about all the constants for a moment, you have a function x^{a}(k-x)^{b}, the way to look at this is differentiate it ans et it to zero, do this for the simple function and see what you get.
 
hunt_mat said:
Forgetting about all the constants for a moment, you have a function x^{a}(k-x)^{b}, the way to look at this is differentiate it ans et it to zero, do this for the simple function and see what you get.

I had done it before. it is (\frac{k l_1}{l_1+l_2})^{l_1} (\frac{k l_2}{l_1+l_2})^{l_2}.but the problem is that each term of the summation is maximized in different values of the given interval as l_1, l_2 vary.
 

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