Solve a_n = 3a_(n-1) + 2, a_0 = 2

Step by step

The question

Question: solve aₙ = 3aₙ₋₁ + 2, starting with a₀ = 2.

Solve means: find a formula for aₙ, so we can get any term without the ones before.

1 First terms

First, a few terms.

Each one comes from the one before.

We start with a₀ = 2.

Then 3 times 2, plus 2.

That is 8.

And so on.

2 The number that stays the same

Is there a number that the rule keeps the same?

Call it c.

If a term were c, the next would be 3c + 2, and it must be c again.

Careful!

c is on both sides.

We solve for it like any equation.

Take away 3c on both sides.

c minus 3c is −2c.

Divide by −2: c = −1.

Check it: 3 times −1, plus 2 is −1 again.

3 The distance from it

Now the distance from c: each term plus 1.

Look at the new column: 3, 9, 27, 81, 243.

Each one is 3 times the one before.

The distance from c grows by times 3.

4 The formula

The distance starts at 3 and is multiplied by 3 each step: 3 times 3 to the n.

Take away 1 on both sides.

3 is one more 3, so the powers join: 3 times 3 to the n is 3 to the n + 1.

5 Check

Check n = 3 with the formula: it gives 80.

The table says 80 too.

Right!

Let's look back.

First: first terms.

Then: the number that stays the same.

Then: the distance from it.

Then: the formula.

Last: check.