Maximum Values for N-Choose-K Relations

[buckleybees]

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?

 

[daon]

The maximum value of "n choose k" happens when k is half of n, or more precisely: if n is even k=n/2, if n is odd then either k=(n-1)/2 or k=(n+1)/2. The choose function is monotonic increasing on the first half of the "k's" and symmetrically decreasing for the latter half.

I'd start by analyzing these in their definitive values:

\(\displaystyle f(k)=\frac{100!2^k}{(100-k)!k!}\)

Note that the function is strictly increasing AT LEAST for \(\displaystyle 0 \le k \le 50\) as per my above explanation. What you might want to do is analyize the assumption that \(\displaystyle \frac{f(k)}{f(k+1)} \le 1\) and look for a k which satisfies its negation.