19 Partial sums of geometric series \( 9 \mathrm{JU} \)
Find the fourth partial sum of the series.
\[
\frac{5}{3}+\frac{25}{9}+\frac{125}{27}+\frac{625}{81}+\frac{3125}{243}+\frac{15625}{729}+\cdots
\]
Write your answer as an integer or a fraction in simplest form.
\[
S_{4}=
\]

Answer

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Answer

Calculate \(S_4 = \frac{\frac{5}{3}\left(\left(\frac{5}{3}\right)^4 - 1\right)}{\frac{5}{3}-1}\).

Steps

Step 1 ：Identify the first term \(a = \frac{5}{3}\), the ratio \(r = \frac{5}{3}\), and the number of terms \(n = 4\).

Step 2 ：Use the geometric series sum formula: \(S_n = \frac{a(r^n - 1)}{r-1}\).

Step 3 ：Calculate \(S_4 = \frac{\frac{5}{3}\left(\left(\frac{5}{3}\right)^4 - 1\right)}{\frac{5}{3}-1}\).