Equivalence of two binomial type equations Given that
$$A=\sum_{i=k}^{2k-1}\binom {2k-1} ix^i(1-x)^{2k-1-i}$$
and
$$B=\sum_{i=k+1}^{2k}\binom {2k} i x^i(1-x)^{2k-i}+\frac{1}{2}\binom {2k} k x^k(1-x)^k$$
I would like to prove that $A=B$
Are there some identities that could be used?
Thanks alot in advance.
 A: Hint: Note that $\binom{2k}{k} = 2\binom{2k-1}{k}$, and show that $B=xA+(1-x)A$.
Okay, long answer, since you are struggling, and I got a part of a hint in comments backwards.
$$\begin{align}xA &= \sum_{i=k}^{2k-1} \binom{2k-1}{i} x^{i+1}(1-x)^{2k-1-i}\\
&= \sum_{j=k+1}^{2k} \binom{2k-1}{j-1} x^{j}(1-x)^{2k-j}\\
&= \left(\sum_{j=k+1}^{2k-1}\binom{2k-1}{j-1} x^{j}(1-x)^{2k-j} \right)+ x^{2k}\end{align}$$
Using substitution $j=i+1$ for the first step, and breaking out the term $j=2k$ in the second step.
On the other hand:
$$\begin{align}(1-x)A &= \sum_{i=k}^{2k-1}\binom{2k-1}{i}x^i(1-x)^{2k-i}\\
&=\binom{2k-1}{k}x^k(1-x)^k +\sum_{i=k+1}^{2k-1} \binom {2k-1}{i}x^i(1-x)^{2k-i}
\end{align}$$
Here, we've just broken out the $i=k$ term.
Now the two $\sum$ terms for $xA$ and $(1-x)A$ have the same range and the same powers of $x$ and $1-x$, so it is easy to add them, and you get:
$$\begin{align}(1-x)A + xA = &\binom{2k-1}{k}x^k(1-x)^k + \\
&\sum_{i=k+1}^{2k-1}\left(\binom{2k-1}{i-1}+\binom{2k-1}{i}\right)x^{i}(1-x)^{2k-i} +\\
&x^{2k}
\end{align}$$
You are almost done. Note $x^{2k}= \binom{2k}{2k} x^{2k}(1-x)^{2k-2k}$, $\binom{2k-1}{k}=\frac{1}{2}\binom{2k}{k}$, and $\binom{2k-1}{i}+\binom{2k-1}{i-1}=\binom{2k}{i}$.
