Let's rewrite and correct the calculation steps clearly:
Given:
[tex]\[ a_1 = 3 \][/tex]
[tex]\[ a_n = 3a_{n-1} + 1 \quad \text{for all } n > 1 \][/tex]
We need to find the values of [tex]\(a_2\)[/tex], [tex]\(a_3\)[/tex], and [tex]\(a_4\)[/tex].
1. Finding [tex]\(a_2\)[/tex]:
[tex]\[
a_2 = 3a_1 + 1
\][/tex]
Substitute [tex]\(a_1 = 3\)[/tex]:
[tex]\[
a_2 = 3 \times 3 + 1 = 9 + 1 = 10
\][/tex]
2. Finding [tex]\(a_3\)[/tex]:
[tex]\[
a_3 = 3a_2 + 1
\][/tex]
Substitute [tex]\(a_2 = 10\)[/tex]:
[tex]\[
a_3 = 3 \times 10 + 1 = 30 + 1 = 31
\][/tex]
3. Finding [tex]\(a_4\)[/tex]:
[tex]\[
a_4 = 3a_3 + 1
\][/tex]
Substitute [tex]\(a_3 = 31\)[/tex]:
[tex]\[
a_4 = 3 \times 31 + 1 = 93 + 1 = 94
\][/tex]
So, the values are:
[tex]\[ a_2 = 10 \][/tex]
[tex]\[ a_3 = 31 \][/tex]
[tex]\[ a_4 = 94 \][/tex]
