You probably know that for [tex]|r|<1[/tex], we have
[tex]\displaystyle\sum_{i=0}^\infty ar^i=\dfrac a{1-r}[/tex]
Your sum starts at [tex]i=1[/tex], so it's equivalent to
[tex]\displaystyle\sum_{i=1}^\infty 12\left(\dfrac12\right)^i=\sum_{i=0}^\infty 12\left(\dfrac12\right)^i-12\left(\dfrac12\right)^0=\dfrac{12}{1-\frac12}-12=12[/tex]
so the answer is A.
