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 $\frac{2^{-a-b+1}(1+x)^{a+b-1}}{1-x}$

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.
