Proof of identity related to sum of binomial coefficients and powers of $2$ Given two positive integers $a,b$, prove that $\left(\sum_{n=a}^{a+b-1}\binom{n-1}{a-1}2^{-n}\right)+\left(\sum_{n=b}^{a+b-1}\binom{n-1}{b-1}2^{-n}\right)=1$
The context comes from this MSE problem, where I indirectly showed that this is true with a combinatorial argument for the specific case of $a=3$ and $b=4$. This is because the first sum computes the probability of $A$ winning and the second sum computes the probability of $B$ winning. Since these are the only $2$ possible events, their sum must be $1$. This argument can be easily extended to a more general case.
While I understand the combinatorial argument, I would like to see an algebraic method. I'm not sure where to start for it.
Also, from the linked post, I (from the other answer) found that $\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n}=2^{-a-b+1}\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}$. Proving this would readily prove the original identity, but I'm not sure how to prove this equation.
I'm looking for something like a generating function proof. I found that $\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n}$ is the coefficient of $x^{b-1}$ in $\frac{1}{(2-x)^a(1-x)}$ and $2^{-a-b+1}\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}$ is the coefficient of $x^{b-1}$ in $\frac{2^{-a-b+1}(1+x)^{a+b-1}}{1-x}$
 A: We prove that $$\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n} = 2^{-a-b+1} \sum_{n=a}^{a+b-1} \binom{a+b-1}{n}$$ for all positive integers $a$ and $b$ by inducting on $b$.
When $b = 1$, the LHS and RHS evaluate to $2^{-a}$, so the statement holds in this case.
We proceed with the induction step. Using the induction hypothesis,
\begin{align*}
\sum_{n=a}^{a+b} \binom{n-1}{a-1}2^{-n} 
&= \left(\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n}\right) + \binom{a+b-1}{a-1}2^{-a-b} \\
&= \left(2^{-a-b+1}\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}\right) + \binom{a+b-1}{a-1}2^{-a-b} \\
&= 2^{-a-b} \left[\left(2\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}\right) + \binom{a+b-1}{a-1} \right].
\end{align*}
Observe the terms in the brackets. Recall that $$\binom{m-1}{k-1} + \binom{m-1}{k} = \binom{m}{k}$$ for all integers $m \geq k \geq 1$. So, by cleverly rearranging the terms in the summation, we have
\begin{align*}
& \:\: \left(2\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}\right) + \binom{a+b-1}{a-1} \\
= & \:\: \binom{a+b-1}{a+b-1} + \sum_{n=a}^{a+b-1}\left(\binom{a+b-1}{n-1} + \binom{a+b-1}{n}\right) \\
= & \: \: 1 + \sum_{n=a}^{a+b-1} \binom{a+b}{n} \\
= & \: \: \sum_{n=a}^{a+b} \binom{a+b}{n}.
\end{align*}
Substituting this back, we get $$\sum_{n=a}^{a+b} \binom{n-1}{a-1}2^{-n} = 2^{-a-b} \sum_{n=a}^{a+b} \binom{a+b}{n}.$$ This completes the induction proof.
A: The first term is
$$\sum_{n=a}^{a+b-1} {n-1\choose a-1} 2^{-n}
= 2^{-a} \sum_{n=0}^{b-1} {n+a-1\choose a-1} 2^{-n}
\\ = 2^{-a} [z^{b-1}] \frac{1}{1-z} \frac{1}{(1-z/2)^a}
= (-1)^a \mathrm{Res}_{z=0}
\frac{1}{z^b} \frac{1}{1-z} \frac{1}{(z-2)^a}.$$
Now residues sum to zero and the residue at infinity is zero by
inspection so to evaluate the residue at zero we require the residues at
one and at  two. For the residue at one we get
$$- (-1)^a \times \frac{1}{(1-2)^a} = -1.$$
For the residue at two we write
$$(-1)^a \mathrm{Res}_{z=2}
\frac{1}{(2+(z-2))^b} \frac{1}{-1-(z-2)} \frac{1}{(z-2)^a}
\\ = - 2^{-b} (-1)^a \mathrm{Res}_{z=2}
\frac{1}{(1+(z-2)/2)^b} \frac{1}{1+(z-2)} \frac{1}{(z-2)^a}
\\ = - 2^{-b} (-1)^a
\sum_{n=0}^{a-1} {n+b-1\choose b-1} (-1)^n 2^{-n}
(-1)^{a-1-n}
\\ = 2^{-b}
\sum_{n=0}^{a-1} {n+b-1\choose b-1} 2^{-n} 
= \sum_{n=b}^{a+b-1} {n-1\choose b-1} 2^{-n}.$$
We have shown that
$$\sum_{n=a}^{a+b-1} {n-1\choose a-1} 2^{-n}
- 1 + \sum_{n=b}^{a+b-1} {n-1\choose b-1} 2^{-n} = 0$$
which is the claim.
