Prove that for all $n \in \mathbb{N}$, $\sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$ $$  \sum_{k=0}^{n-1}{n+k-1\choose k}\frac 1{2^{n+k}}=\frac12$$
To be honest, I can't really get started,
I would like to ask everyone to give me an idea of ​​how to solve it, give me a starting push, thank you.
 A: By the way, the product being summed looks like the PMF of the negative binomial distribution, that to flip a fair coin repeatedly until having $n$ heads, there's $\frac 12$ probability that there were at most $n-1$ tails before the $n$th head.
Let $\Pr(A)$ be the probability that there were at most $n-1$ tails before the $n$th head.
Then $1-\Pr(A)$ would be the probability that the $n$th tail appears while there were at most $n-1$ heads.
For a fair coin, head and tail are symmetric, and so
$$\Pr(A) = 1-\Pr(A) = \frac 12$$
A: we rewrite the sum like this ;$$\sum_{k=0}^{k=n-1}\binom{n+k-1}{n-1}\frac{1}{2^{n+k}}$$
this is just the coefficient of $x^{n-1}$ in expansion $$\frac{{(1+x)}^{n-1}}{2^n}+\frac{{(1+x)}^n}{2^{n+1}}...+\frac{{(1+x)}^{2n-2}}{2^{2n-1}}$$
Recognise this as a GP: hence we seek the coefficient of $x^{n-1}$ in $$\frac{1}{2^{n-1}}{(1+x)}^{n-1}\frac{1-{(\frac{x+1}{2})}^n}{1-x}$$
or $$\frac{{(1+x)}^{n-1}(1+x+x^2..)-{(1+x)}^{2n-1}(1+x+x^2+x^3....)}{2^{2n-1}}$$ The coefficent is $$\frac{\left(\binom{n-1}{0}+\binom{n-1}{1}..+\binom{n-1}{n-1}\right)-\left(\binom{2n-1}{0}+\binom{2n-1}{1}+...\binom{2n-1}{n-1}\right)}{2^{2n-1}}$$
$$=\frac{2^{2n-1}-\frac{1}{2}2^{2n-1}}{2^{2n-1}}=\frac{1}{2}$$
