How to prove that this binomial sum remains positive for $x>1$?

Question

Let's say you have this function for real numbers $x>1$, for some positive integer $n \geq 1$

$$ \sum_{k=0}^{\left \lfloor n/2 \right \rfloor} {x \choose 2k+\frac{1-(-1)^n}{2}} $$

How would you prove that this function is positive for all $x>1$ and $x \in \mathbb{R}$? For positive integers, the binomial coefficients are positive for some real number $x>n$, but it's not clear for a real number since the binomial coefficient expanded to the real number field requires the use of the gamma function which changes signs depending on the negative real number. I know that for the case of $n$ odd this polynomial has 1 real root which is $x=0$, and for $n$ even it has roots $x=0,1$ but not sure where to go from here to prove this for all real numbers $x>1$?

Answer 1

This is not an answer. since no proof

If $n=2m$ $$S_m=\sum_{k=0}^m \binom{x}{2 k}$$ $$S_m=2^{x-1}-\binom{x}{2 (m+1)} \,\, _3F_2\left(1,\frac{2 m+2-x}{2} ,\frac{2 m+3-x}{2};\frac{2m+3}{2},m+2 ;1\right)$$

$$S_m=1+2^{x-1}+\frac {x(x-1) }{(2m)!}\,P_m(x)-\, _2F_1\left(\frac{1-x}{2},-\frac{x}{2 };\frac{1}{2};1\right)$$ where the first polynomials are $$\left( \begin{array}{cc} m & P_m(x) \\ 1 & 1 \\ 2 & x^2-5 x+18 \\ 3 & x^4-14 x^3+101 x^2-304 x+660 \\ 4 & x^6-27 x^5+351 x^4-2449 x^3+10760 x^2-25052 x+42000 \\ \end{array} \right)$$

Computed for a bunch of values of $m$, the polynomials $P_m(x)$ are all positive and show a minimum value at $x=\frac 52-\epsilon_m$

$$\left( \begin{array}{cc} m & \epsilon_m \\ 2 & 0.0000000 \\ 3 & 0.0148855 \\ 4 & 0.0205105 \\ 5 & 0.0231958 \\ 6 & 0.0246705 \\ 7 & 0.0255607 \\ 8 & 0.0261362 \\ 9 & 0.0265281 \\ \end{array} \right)$$

Morover, since asymptotically

$$\,_2F_1\left(\frac{1-x}{2},-\frac{x}{2};\frac{1}{2};1\right)\sim 2^{x-1}$$ therefore

$$S_m\sim 1+\frac {x(x-1) }{(2m)!}\,P_m(x)$$

Edit

For small values of $x$

$$\,_2F_1\left(\frac{1-x}{2},-\frac{x}{2};\frac{1}{2};1\right)=\sum_{n=0}^\infty \frac{\log(2)^n}{n!}(x-1)^n=2^{x-1}$$