When will $f(i):=\binom{2k-1}{i}\Big((1-p)^i(1+p)^{2k-1-i}-(1+p)^i(1-p)^{2k-1-i} \Big)$ attain maximum? When will 
$$f(i):=\binom{2k-1}{i}\Big((1-p)^i(1+p)^{2k-1-i}-(1+p)^i(1-p)^{2k-1-i} \Big)$$
attain maximum among $i=0,1,\dots,k-1$,
for very large positive integer $k$, and $p\in (0,1)$ with $p=\Omega(1/k^{1/2})$?
Or with the same assumption, can we bound it from above by $2^{2k-1}\exp(-\Omega(k^{1/2}))$?

Note that it is the difference (after factoring $2^{-(2k-1)}$ out)
 $$\mathbb{P}(X=i)-\mathbb{P}(Y=i),$$
where $X$ is binomial random variable $Bin(2k-1,\frac{1-p}{2})$ and $Y$ is binomial random variable $Bin(2k-1,\frac{1+p}{2})$. Therefore $f(i)$ should always be possible  for $i=0,1,\dots,k-1$.
 A: My numerical experiments suggest that the subtrahend is negligible when $f(i)$ attains its maximum. Then we have to consider $i$ at which $\mathbb{P}(X=i)$ attains its maximum (which seems to happen when $i\approx (1-p)(k-1/2)$) and the magnitude of $f(i)$. For this we shall use heuristic approximations, which are not rigorous, but convincing. 
Applying Stirling’s formula, we obtain 
$${2k-1 \choose i}=\frac{(2k-1)!}{i!(2k-1-i)!}\approx$$ $$\frac{\sqrt{2\pi(2k-1)}\left(\frac {2k-1}e \right)^{2k-1}}{\sqrt{2\pi i}\left(\frac ie \right)^i\sqrt{2\pi(2k-1-i)}\left(\frac {2k-1-i}e \right)^{2k-1-i}}=$$ $$\frac{\sqrt{(2k-1)}\left(2k-1\right)^{2k-1}}{\sqrt{2\pi}\sqrt{i(2k-1-i)}i^i\left(2k-1-i \right)^{2k-1-i}}\approx$$ 
$$\frac{\sqrt{(2k-1)}\left(2k-1\right)^{2k-1}}{\sqrt{2\pi}\sqrt{(1-p)(1+p)(k-1/2)^2}((1-p)(k-1/2))^{(1-p)(k-1/2)}\left((1+p)(k-1/2)\right)^{(1+p)(k-1/2)}}=$$ 
$$\frac {2^{2k-1}}{\sqrt{\pi(k-1/2)(1-p^2)}}\cdot (1-p)^{-(1-p)(k-1/2)} (1+p)^{-(1+p)(k-1/2)}=$$
$$\frac {2^{2l}}{\sqrt{\pi l(1-p^2)}}\cdot (1-p)^{-(1-p)l} (1+p)^{-(1+p)l},$$
where $l=k-1/2$.
Put
$$A=(1-p)^i(1+p)^{2k-1-i}\approx (1-p)^{(1-p)l}(1+p)^{(1+p)l}$$ 
$$B=(1+p)^i(1-p)^{2k-1-i}\approx (1+p)^{(1-p)l}(1-p)^{(1+p)l}$$
Then $$\frac BA\approx\left(\frac {1-p}{1+p}\right)^{2pl}$$
which is negligible for fixed $p\in (0,1)$ and big $l$.
Thus 
$$f(i)\approx {2k-1 \choose i} A\approx \frac {2^{2l}}{\sqrt{\pi l(1-p^2)}}\cdot (1-p)^{-(1-p)l} (1+p)^{-(1+p)l}(1-p)^{(1-p)l}(1+p)^{(1+p)l}=$$ $$\frac {2^{2l}}{\sqrt{\pi l(1-p^2)}}$$
That is $\frac {f(i)}{ 2^{2k-1}}$ is only as small as $\frac 1{\sqrt{\pi (k-1/2)(1-p^2)}}$, but not as 
$\exp(-\Omega(k^{1/2}))$.
