Question 758998
The nth term of a geometric series is normally written {{{a_n = a*r^(n-1)}}}
For this series, we know that the infinite sum {{{S = a/(1-r)}}}
The series {{{(1/2)^n}}} has a=1 and r=1/2. This series has terms 1/2, 1/4, 1/8...
The series above has terms 1, 1/2, 1/4, 1/8...
So the two series differ by 1.
The sum of the series {{{a_n = (1/2)^(n-1) = 1/(1-(1/2)) = 2}}}
Therefore the sum of the series {{{a_n = (1/2)^n = 2-1 = 1}}}
I'll leave it to you to figure out the other one.