About the combinatorial question: the proof follows from the identity
$$
{2k-1 \choose i-1} + {2k-1 \choose i} = {2k \choose i}
$$
Thus
$$
\sum_{i=0}^{k-1} {2k\choose i} p^{2k-i} (1-p)^i = 
\sum_{i=0}^{k-1} {2k-1 \choose i} p^{2k-i} (1-p)^i + 
\sum_{i=0}^{k-1} {2k-1 \choose i-1} p^{2k-i} (1-p)^i
$$
which implies
$$
\sum_{i=0}^{k-1} {2k\choose i} p^{2k-i} (1-p)^i = 
\sum_{i=0}^{k-2} {2k-1 \choose i} p^{2k-1-i} (1-p)^i + 
{2k-1 \choose k-1} p^{k+1} (1-p)^{k-1}
$$
and
$$
\frac{1}{2}{2k\choose k} = {2k-1\choose k-1}
$$
implying that
$$
{2k-1 \choose k-1} p^{k+1} (1-p)^{k-1} + \frac{1}{2}{2k\choose k} p^{k} (1-p)^{k} = {2k-1\choose k-1 } p^{k} (1-p)^{k-1}
$$
which establishes your identity.

About your "Edit" question, I think you mean the theorem in complex calculus that states that, if an analytic function is constant over an interval, it is constant everywhere.