Question

suppose that a sequence is defined as follows.
$a_1 = -1,\quad a_n = -3a_{n - 1} + 6\quad\text{for}\quad n \geq 2$
list the first four terms of the sequence.

Explanation:

Step1: Find \(a_2\)

We know \(a_1 = - 1\) and the recurrence relation \(a_{n}=-3a_{n - 1}+6\) for \(n\geq2\). For \(n = 2\), we substitute \(n-1=1\) into the recurrence relation. So \(a_{2}=-3a_{1}+6\). Substitute \(a_{1}=-1\) into the formula: \(a_{2}=-3\times(-1)+6=3 + 6=9\).

Step2: Find \(a_3\)

For \(n = 3\), we use the recurrence relation \(a_{n}=-3a_{n - 1}+6\). Here \(n-1 = 2\), so \(a_{3}=-3a_{2}+6\). We know \(a_{2}=9\) from Step 1, substitute it in: \(a_{3}=-3\times9+6=-27 + 6=-21\).

Step3: Find \(a_4\)

For \(n = 4\), we use the recurrence relation \(a_{n}=-3a_{n - 1}+6\). Here \(n - 1=3\), so \(a_{4}=-3a_{3}+6\). We know \(a_{3}=-21\) from Step 2, substitute it in: \(a_{4}=-3\times(-21)+6 = 63+6=69\).

Answer:

The first four terms of the sequence are \(a_1=-1\), \(a_2 = 9\), \(a_3=-21\), \(a_4 = 69\)