For which $k$ the following equation has the greatest value: $ k \cdot\binom{99}{k} $?

Question

After some manipulation I got $$\frac{99!}{(k-1)!\cdot(99-k)!} $$

So I guess I have to find $k$ for which I get the smallest denominator, but I don't know where to go from there.

Answer 1

HINT:

Note that

$$(k+1)\binom{99}{k+1}-(k)\binom{99}{k}=\frac{99!}{(k-1)!(98-k)!}\left(\frac{99-2k}{k(99-k)}\right)$$

Now determine the values of $k$ for which $k\binom{99}{k}$ is increasing and the values of $k$ for which $k\binom{99}{k}$ are decreasing. Can you conclude now?

Answer 2

Hint:

Set $a_k=k\,\dbinom{99} k$ and show the sequence $(a_k)_{0\le k\le 99}$ is first increasing, then decreasing by comparing the ratio $\;\dfrac{a_{k+1}}{a_k}$ to $1$. This ratio simplifies to a homographic function of $k$.

Answer 3

$$ k\binom{99}k=99\binom{98}{k-1}, $$ so it suffices to maximize $\binom{98}{k-1}$. The largest binomial coefficient is in the middle, so...