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find the first 5 terms.\\begin{cases} a_0 = 81 \\\\ a_n = \\frac{a_{n -…

Question

find the first 5 terms.\

$$\begin{cases} a_0 = 81 \\\\ a_n = \\frac{a_{n - 1}}{3} \\end{cases}$$

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$$\begin{tabular}{|c|c|c|c|c|} \\hline $a_0$ & $a_1$ & $a_2$ & $a_3$ & $a_4$ \\\\ \\hline ? & & & & \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Identify \(a_0\)

Given \(a_0 = 81\).

Step2: Calculate \(a_1\)

Using the formula \(a_n=\frac{a_{n - 1}}{3}\), for \(n = 1\), we have \(a_1=\frac{a_0}{3}\). Substitute \(a_0 = 81\) into the formula: \(a_1=\frac{81}{3}=27\).

Step3: Calculate \(a_2\)

For \(n = 2\), \(a_2=\frac{a_1}{3}\). Substitute \(a_1 = 27\): \(a_2=\frac{27}{3}=9\).

Step4: Calculate \(a_3\)

For \(n = 3\), \(a_3=\frac{a_2}{3}\). Substitute \(a_2 = 9\): \(a_3=\frac{9}{3}=3\).

Step5: Calculate \(a_4\)

For \(n = 4\), \(a_4=\frac{a_3}{3}\). Substitute \(a_3 = 3\): \(a_4=\frac{3}{3}=1\).

Answer:

\(a_0 = 81\), \(a_1 = 27\), \(a_2 = 9\), \(a_3 = 3\), \(a_4 = 1\)