# Maximum Values for N-Choose-K Relations

• Mar 10th 2010, 12:25 PM
buckleybees
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?
• Mar 11th 2010, 01:07 AM
Opalg
Quote:

Originally Posted by 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?

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.