A geometric series:
\(\sum^{\infty}_{i=1}=a_1 \times r^{i-1}\)
It's convergent if |r|<1.
It's divergent if |r|≥1.
The sum can be found if it's a convergent series; it's equal to \(\frac{a_1}{1-r}\).
3.
\(\sum^{\infty}_{i=1} 12 (\frac{3}{5})^{i-1} \\ \\
a_1=12 \\
r=\frac{3}{5} \\ \\
|r|<1 \hbox{ so it's convergent} \\ \\
\sum^{\infty}_{i=1} 12 (\frac{3}{5})^{i-1}=\frac{12}{1-\frac{3}{5}}=\frac{12}{\frac{5}{5}-\frac{3}{5}}=\frac{12}{\frac{2}{5}}=12 \times \frac{5}{2}=6 \times 5=30\)
The answer is: This is a convergent geometric series. The sum is 30.
4.
\(\sum^{\infty}_{i=1} 15(4)^{i-1} \\ \\ a_1=15 \\ r=4 \\ \\ |r| \geq 1 \hbox{ so it's divergent}\)
The answer is: This is a divergent geometric series. The sum cannot be found.
