No closed form for the partial sum of ${n\choose k}$ for $k \le K$? In Concrete Mathematics, the authors state that there is no closed form for
$$\sum_{k\le K}{n\choose k}.$$
This is stated shortly after the statement of (5.17) in section 5.1 (2nd edition of the book).
How do they know this is true?

Proof of the answer by nczksv.
$$
\begin{array}{l}
f = \dfrac{(1+x)^n}{x^K \cdot (1-x)} 
= \left(\dfrac{1+x}{x}\right)^n \cdot \dfrac{x^{n-K}}{1-x}\\
= \left(4^n+1\right)^n \cdot 4^n \cdot \dfrac{x^{n-K}}{4^n - 1}\\
= \dfrac{\left(4^n+1\right)^n \cdot 4^n }{(4^n - 1)(4^n)^{(n-K)}}
\end{array}
$$
Let $A = 4^n - 1$, mod = $A+1$. Consider both flooring function and mod function, 
Note that 
$$\forall r > 1, ~ 
\left\lfloor
\frac{(A+1)^r}{A} \right\rfloor
 \equiv \left\lfloor (A+1)^{r-1} + \frac{(A+1)^{r-1}}{A}  \right\rfloor
= \left\lfloor \frac{(A+1)^{r-1}}{A} \right\rfloor
= \frac{A+1}{A}$$
we have 
$$
\begin{array}{l}
f = \dfrac{\left(A+2\right)^n }{A \cdot (A+1)^{n-K-1}} \\
= \dfrac{ \sum_{k=0}^{K} {n \choose k} (A+1)^{n-k} + \sum_{k=K+1}^{n} {n \choose k} (A+1)^{n-k} }{A \cdot (A+1)^{n-K-1}}\\
= \dfrac{ \sum_{k=0}^{K} {n \choose k} (A+1)^{K+1-k} }{A} + 
\dfrac{ \sum_{k=K+1}^{n} {n \choose k} (A+1)^{K+1-k} }{A}\\
= \sum_{k=0}^{K} {n \choose k} +  \dfrac{ \sum_{k=0}^{K} {n \choose k} }{A} + 
\dfrac{ \sum_{k=K+1}^{n} {n \choose k} (A+1)^{K+1-k} }{A}
= \sum_{k=0}^{K} {n \choose k} + 0
\end{array}
$$
