Determine $\frac{f''(\frac{1}{2})}{f'(\frac{1}{2})}$ if $f(x) = \sum_{k=0}^{1000} \ {2015 \choose k}\ x^k(1-x)^{2015-k}$ 
Problem : Determine $\frac{f''(\frac{1}{2})}{f'(\frac{1}{2})}$ if $f(x) = \sum_{k=0}^{1000} \ {2015 \choose k}\ x^k(1-x)^{2015-k}$

Trying to simply brute force the problem, yields the following derivatives :
$$
\begin{equation}
\begin{split}
f(x) &= \sum_{k=0}^{1000} \ {2015 \choose k}\ x^k(1-x)^{2015-k} \\
f'(x) &= \sum_{k=0}^{1000} \ {2015 \choose k}\ \left[(kx^{k-1})(1-x)^{2015-k} - x^k(2015-k)(1-x)^{2014}\right]\\
f''(x) &= \sum_{k=0}^{1000} \ {2015 \choose k}\ \left[(k(k-1)x^{k-2})\right(1-x)^{2015-k} - 2kx^{k-1}(2015-k)(1-x)^{2014-k}  + x^k(1-x)^{2014-k}\left[1+ (2015-k)(2014-k)(1-x)\right]]\\
\end{split}
\end{equation}
$$
But trying to evaluate $\frac{f''(\frac{1}{2})}{f'(\frac{1}{2})}$, given the above derivatives for $f'(x)$ and $f''(x)$ by conventional means is next to impossible. 
Is there a simpler way to find a solution to this problem, or a method/trick to simplify the derivatives further to get them into a form that can be evaluated?
 A: Define $$f_{m,n}(x) = \sum_{k=0}^m \binom{n}{k} x^k (1-x)^{n-k}, \quad 0 < m \le n, \quad 0 < x < 1.$$  Then 
$$f_{m,n}'(x) = \sum_{k=0}^m \binom{n}{k} k x^{k-1} (1-x)^{n-k} - (n-k) \binom{n}{k} x^k (1-x)^{n-k-1}.$$
Using the identities 
$$k \binom{n}{k} = n \binom{n-1}{k-1}, \quad (n-k)\binom{n}{k} = n \binom{n-1}{k},$$ 
we obtain 
$$\begin{align*} f_{m,n}'(x) &= n \sum_{k=1}^{m} \binom{n-1}{k-1} x^{k-1} (1-x)^{n-k} - n \sum_{k=0}^m \binom{n-1}{k} x^k (1-x)^{n-k-1} \\ 
&= n \sum_{k=0}^{m-1} \binom{n-1}{k} x^k (1-x)^{n-k-1} - n \sum_{k=0}^m \binom{n-1}{k} x^k (1-x)^{n-k-1} \\ 
&= -n \binom{n-1}{m} x^m (1-x)^{n-m-1}, 
\end{align*}$$ 
whenever such an expression is defined.  Consequently, 
$$\begin{align*} \frac{f_{m,n}''(x)}{f_{m,n}'(x)} &= \frac{d}{dx}\left[\log f_{m,n}'(x) \right] \\ 
&= \frac{d}{dx} \left[ \log \left(-n \binom{n-1}{m} \right) + m \log x + (n-m-1) \log (1-x) \right] \\ 
&= \frac{m}{x} - \frac{n-m-1}{1-x}. 
\end{align*}$$
For $x = 1/2$, we obtain the special case 
$$\frac{f_{m,n}''(1/2)}{f_{m,n}'(1/2)} = 2(1+2m-n),$$ 
and for the case $n = 2015$, $m = 1000$, we obtain the answer of $-28$.
A: More generally, let $$f(x) = \sum_{k=0}^K {N \choose k} x^k (1-x)^{N-k}$$
This is $\mathbb P(B \le K)$ where $B$ is a binomial random variable with parameters $N$ and $x$. It can be expressed in terms of a hypergeometric function.
With Maple's help, I get a simplification
$$ \dfrac{f''(1/2)}{f'(1/2)} = -2 N+4 K +2$$
Hmmm, there's got to be a good reason for this...
