Question

suppose that a sequence is defined as follows. $a_1 = 5$, $a_n = -2a_{n - 1} + 6$ for $n geq 2$ list the first four terms of the sequence.

Explanation:

Step1: Find the second term (\(a_2\))

We know \(a_1 = 5\) and the recurrence relation \(a_n=-2a_{n - 1}+6\) for \(n\geq2\). For \(n = 2\), we substitute \(n-1 = 1\) into the recurrence relation. So \(a_2=-2a_1 + 6\). Substituting \(a_1 = 5\), we get \(a_2=-2\times5+6=-10 + 6=-4\).

Step2: Find the third term (\(a_3\))

For \(n = 3\), we use the recurrence relation \(a_n=-2a_{n - 1}+6\) with \(n-1 = 2\). So \(a_3=-2a_2+6\). We already found \(a_2=-4\), so substituting that in, we get \(a_3=-2\times(-4)+6 = 8 + 6=14\).

Step3: Find the fourth term (\(a_4\))

For \(n = 4\), we use the recurrence relation \(a_n=-2a_{n - 1}+6\) with \(n-1 = 3\). So \(a_4=-2a_3+6\). We found \(a_3 = 14\), so substituting that in, we get \(a_4=-2\times14+6=-28 + 6=-22\).

Answer:

The first four terms are \(a_1 = 5\), \(a_2=-4\), \(a_3 = 14\), \(a_4=-22\)