Solve a_n = 3a_(n-1) - 4, a_0 = 5 Step by Step

Step by step

The question

Question: solve aₙ = 3aₙ₋₁ − 4, starting with a₀ = 5.

a_n = 3a_{n − 1} − 4, a₀ = 5

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 a₀ = 5.

start

Then 3 times 5, minus 4.

3 × 5 − 4

That is 11.

And so on.

3 × 11 − 4

3 × 29 − 4

3 × 83 − 4

245

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 − 4, and it must be c again.

c = 3c − 4

Careful!

c is on both sides.

We solve for it like any equation.

Careful: not c = 3 − 4 = -1 ✗

Take away 3c on both sides.

c − 3c = −4

c minus 3c is −2c.

−2c = −4

Divide by −2: c = 2.

c = 2

Check it: 3 times 2, minus 4 is 2 again.

check: 3 × 2 − 4 = 2 ✓

3 The distance from it

Now the distance from c: each term minus 2.

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.

a_n − 2 = 3 × 3ⁿ

Add 2 on both sides.

a_n = 3 × 3ⁿ + 2

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

= 3^{n + 1} + 2

5 Check

Check n = 3 with the formula: it gives 83.

n = 3: formula gives 83

The table says 83 too.

Right!

table: a₃ = 83

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