Question

a1=32, r=-0.06 Find the sum of the given infinite geometric series. I hate sums :(

Answer 1

You're given the infinite geometric series
\[\sum_{n=0}^\infty32\left(-0.06\right)^n\\
\sum_{n=0}^\infty32\left(-\frac{3}{50}\right)^n\]
The sum of an infinite geometric series is
\[\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\]