how to simplify the series I need to find the value of
$\sum_{w=1} ^ {n-1} { n \choose w} 0.5^w (1-0.5)^{n-w}$
I know I should get $1-2(0.5)^n$
But I dont understand how to get there
You can look at the key to the soution in $2$ ways. The first is through the binomial distribution and probability, and the second is the binomal theorem.
1.
The general probability of the discrete random variable, $X$, modelled by a binomial distribution, is given by
$$P(X=r)=\binom{n}{r}(p)^r(1-p)^{n-r}$$
where $r$ is the number of successful outcomes, $n$ is the number of trials and $p$ is the probability of success. As all probabilities for a given event add up to $1$, we have
$$\sum_{r=0}^{n}\binom{n}{r}(p)^r(1-p)^{n-r}=1$$
2.
Alternatively, we have the binomial theorem which states that
$$(q+p)^n=\binom{n}{0}(q)^n(p)^0+\binom{n}{1}(q)^{n-1}(p)^1+...+\binom{n}{r}(q)^{n-r}(p)^r+...+\binom{n}{n}(q)^0(p)^n=\sum_{r=0}^{n}\binom{n}{r}(q)^{n-r}(p)^r$$
Now, if we substitute $q=1-p$ we obtain the following:
$$(1-p+p)^n=1=\sum_{r=0}^{n}\binom{n}{r}(1-p)^{n-r}(p)^r=\sum_{r=0}^{n}\binom{n}{r}(p)^r(1-p)^{n-r}$$

Either way, I think the answer to your question is
$$\sum_{r=1}^{n-1}\binom{n}{r}(p)^r(1-p)^{n-r}=(\sum_{r=0}^{n}\binom{n}{r}(p)^r(1-p)^{n-r})-\binom{n}{0}(p)^0(1-p)^n-\binom{n}{n}(p)^n(1-p)^0=1-\binom{n}{0}(p)^0(1-p)^n-\binom{n}{n}(p)^n(1-p)^0=1-(1-p)^n-(p)^n$$
Substitute $p=0.5$, as it is in your question, and we find that it is equal to
$$1-0.5^n-0.5^n=1-2(0.5)^n$$
as required. I hope that helps :))
(P.S: You could simplify this slightly more by writing $1-(0.5)^{n-1}$.)
You can look at the key to the soution in $2$ ways. The first is through the binomial distribution and probability, and the second is the binomal theorem.
1.
The general probability of the discrete random variable, $X$, modelled by a binomial distribution, is given by
$$P(X=r)=\binom{n}{r}(p)^r(1-p)^{n-r}$$
where $r$ is the number of successful outcomes, $n$ is the number of trials and $p$ is the probability of success. As all probabilities for a given event add up to $1$, we have
$$\sum_{r=0}^{n}\binom{n}{r}(p)^r(1-p)^{n-r}=1$$
2.
Alternatively, we have the binomial theorem which states that
$$(q+p)^n=\binom{n}{0}(q)^n(p)^0+\binom{n}{1}(q)^{n-1}(p)^1+...+\binom{n}{r}(q)^{n-r}(p)^r+...+\binom{n}{n}(q)^0(p)^n=\sum_{r=0}^{n}\binom{n}{r}(q)^{n-r}(p)^r$$
Now, if we substitute $q=1-p$ we obtain the following:
$$(1-p+p)^n=1=\sum_{r=0}^{n}\binom{n}{r}(1-p)^{n-r}(p)^r=\sum_{r=0}^{n}\binom{n}{r}(p)^r(1-p)^{n-r}$$

Either way, I think the answer to your question is
$$\sum_{r=1}^{n-1}\binom{n}{r}(p)^r(1-p)^{n-r}=(\sum_{r=0}^{n}\binom{n}{r}(p)^r(1-p)^{n-r})-\binom{n}{0}(p)^0(1-p)^n-\binom{n}{n}(p)^n(1-p)^0=1-\binom{n}{0}(p)^0(1-p)^n-\binom{n}{n}(p)^n(1-p)^0=1-(1-p)^n-(p)^n$$
Substitute $p=0.5$, as it is in your question, and we find that it is equal to
$$1-0.5^n-0.5^n=1-2(0.5)^n$$
as required. I hope that helps :))
(P.S: You could simplify this slightly more by writing $1-(0.5)^{n-1}$.)