the nth term is
an=a1(r)^(n-1)
an=1(2)^(n-1)
a1=1
r=2
the sum of a geometric seequence is
\(S_{n}=\frac{a_{1}(1-r^{n})}{1-r}\)
a1=1
r=2
we want to find \(S_{10}\) (since we minus 1, the highest exponet is 9 so add 1 to make it correct)
\(S_{10}=\frac{1(1-2^{10})}{1-2}\)
\(S_{10}=\frac{(1-1024}{-1}\)
\(S_{10}=\frac{(-1023}{-1}\)
\(S_{10}=1023\)
