Solve in Terms of the Arbitrary Variable a a_n=3a_n-1+2

$a_{n} = 3 a_{n - 1} + 2$

Step 1

Rewrite the equation as $3 a_{n - 1} + 2 = a_{n}$ .

$3 a_{n - 1} + 2 = a_{n}$

Step 2

Move all terms not containing $a「+n-1」$ to the right side of the equation.

Subtract $3 a_{n - 1}$ from both sides of the equation.

$2 = a_{n} - 3 a_{n - 1}$

Subtract $2$ from both sides of the equation.

$0 = a_{n} - 3 a_{n - 1} - 2$

Step 3

Rewrite the equation as $a_{n} - 3 a_{n - 1} - 2 = 0$ .

$a_{n} - 3 a_{n - 1} - 2 = 0$