Question

What is the sum of this infinite geometric series

Answer 1

You probably know that for
`|r|<1`, we have

`\displaystyle\sum_(i=0)^\infty ar^i=\frac a{1-r}`

Your sum starts at
`i=1`, so it's equivalent to

`\displaystyle\sum_(i=1)^\infty 12\left(\frac12\right)^i=\sum_(i=0)^\infty 12\left(\frac12\right)^i-12\left(\frac12\right)^0=(12)/(1-\frac12)-12=12`

so the answer is A.