Solve a_n = 3a_(n-1) - 2, a_1 = 4 Step by Step

Step by step

The question

Question: solve aₙ = 3aₙ₋₁ − 2, starting with a₁ = 4.

a_n = 3a_{n − 1} − 2, a₁ = 4

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

solve = a formula for a_n

1 First terms

First, a few terms.

Each one comes from the one before.

We start with a₁ = 4.

start

Then 3 times 4, minus 2.

3 × 4 − 2

That is 10.

And so on.

3 × 10 − 2

3 × 28 − 2

3 × 82 − 2

244

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.

c = 3c − 2

Careful!

c is on both sides.

We solve for it like any equation.

Careful: not c = 3 − 2 = 1 ✗

Take away 3c on both sides.

c − 3c = −2

c minus 3c is −2c.

−2c = −2

Divide by −2: c = 1.

c = 1

Check it: 3 times 1, minus 2 is 1 again.

check: 3 × 1 − 2 = 1 ✓

3 The distance from it

Now the distance from c: each term minus 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.

× 3 each time

4 The formula

The distance starts at 3 and is multiplied by 3 each step: 3 times 3 to the n − 1.

a_n − 1 = 3 × 3^{n − 1}

Add 1 on both sides.

a_n = 3 × 3^{n − 1} + 1

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

= 3ⁿ + 1

5 Check

Check n = 4 with the formula: it gives 82.

n = 4: formula gives 82

The table says 82 too.

Right!

table: a₄ = 82

Let's look back.

First: first terms.

Then: the number that stays the same.

Then: the distance from it.

Then: the formula.

Last: check.

Answer