Consider the [tex]n[/tex]-th partial sum of a geometric series with first term [tex]a[/tex] and common ratio [tex]r[/tex] between consecutive terms. Then the sum of the first [tex]n[/tex] terms of this series is
[tex]S = a + ar + ar^2 + \cdots + ar^{n-1}[/tex]
Multiply both sides by [tex]r[/tex], then subtract [tex]rS[/tex] from [tex]S[/tex] and solve for [tex]S[/tex].
[tex]rS = ar + ar^2 + ar^3 + \cdots + ar^n[/tex]
[tex]S - rS = a - ar^n[/tex]
[tex]\implies S = \dfrac{a(1-r^n)}{1-r}[/tex]
In the given series, we have [tex]a=1[/tex] and [tex]r=2[/tex]; then the sum of the series with [tex]n=8[/tex] terms is
[tex]S = \dfrac{1-2^8}{1-2} = 2^8 - 1 = \boxed{255}[/tex]
