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find the first 5 terms. \\begin{cases} a_0 = 3 \\\\ a_n = 2a_{n - 1} \\…

Question

find the first 5 terms.
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$$\begin{cases} a_0 = 3 \\\\ a_n = 2a_{n - 1} \\end{cases}$$

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

Explanation:

Step1: Identify \( a_0 \)

Given \( a_0 = 3 \).

Step2: Calculate \( a_1 \)

Use the recurrence relation \( a_n = 2a_{n - 1} \). For \( n = 1 \), \( a_1 = 2a_0 \). Substitute \( a_0 = 3 \), so \( a_1 = 2\times3 = 6 \).

Step3: Calculate \( a_2 \)

For \( n = 2 \), \( a_2 = 2a_1 \). Substitute \( a_1 = 6 \), so \( a_2 = 2\times6 = 12 \).

Step4: Calculate \( a_3 \)

For \( n = 3 \), \( a_3 = 2a_2 \). Substitute \( a_2 = 12 \), so \( a_3 = 2\times12 = 24 \).

Step5: Calculate \( a_4 \)

For \( n = 4 \), \( a_4 = 2a_3 \). Substitute \( a_3 = 24 \), so \( a_4 = 2\times24 = 48 \).

Answer:

\( a_0 = 3 \), \( a_1 = 6 \), \( a_2 = 12 \), \( a_3 = 24 \), \( a_4 = 48 \)