Question

write the first four terms of the sequence defined by $a_n = \

$$\begin{cases} 5, & n = 1 \\\\ a_{n-1} + 3, & n > 1 \\end{cases}$$

$.
\\(\boldsymbol{\circ}\\) 5, 4, 3, 2, ...
\\(\boldsymbol{\circ}\\) 5, 8, 11, 14, ...
\\(\boldsymbol{\circ}\\) 5, 15, 45, 135, ...
\\(\boldsymbol{\circ}\\) 5, 2, -1, -4, ...

Explanation:

Step1: Find \(a_1\)

Given the sequence definition, when \(n = 1\), \(a_1 = 5\).

Step2: Find \(a_2\)

For \(n>1\), the formula is \(a_n=a_{n - 1}+3\). When \(n = 2\), \(a_2=a_{2 - 1}+3=a_1 + 3\). Substitute \(a_1 = 5\), we get \(a_2=5 + 3=8\).

Step3: Find \(a_3\)

When \(n = 3\), \(a_3=a_{3 - 1}+3=a_2 + 3\). Substitute \(a_2 = 8\), we get \(a_3=8 + 3=11\).

Step4: Find \(a_4\)

When \(n = 4\), \(a_4=a_{4 - 1}+3=a_3 + 3\). Substitute \(a_3 = 11\), we get \(a_4=11 + 3=14\).

Answer:

The first four terms of the sequence are \(5, 8, 11, 14,\dots\) (the second option: 5, 8, 11, 14, …)