Given:
The geometric series
[tex]\sum_{n\mathop{=}1}^9500(1.05)^{n-1}[/tex]
Required:
What is the sum of gemetric series?
Explanation:
The formula for sum of geometric series with finite terms:
[tex]\begin{gathered} S_n=\frac{a(r^n-1)}{r-1} \\ Where, \\ a(first\text{ }term) \\ r(common\text{ }ratio) \end{gathered}[/tex]
So, the series:
[tex]\begin{gathered} =500+500(1.05)+500(1.05)^2+...... \\ So,r=\frac{500(1.05)}{500} \\ r=1.05 \end{gathered}[/tex]
Now, sum is:
[tex]\begin{gathered} S_9=\frac{500(1.05^9-1)}{(9-1)} \\ =\frac{500(0.551)}{8} \\ =\frac{275.5}{8} \\ =34.4 \end{gathered}[/tex]
Answer:
The sum is 34.4
