Solve a_n = 3a_(n-1) + 4a_(n-2), a_0 = 2, a_1 = 3

Step by step

The question

Question: solve aₙ = 3aₙ₋₁ + 4aₙ₋₂, with a₀ = 2 and a₁ = 3.

1 Everything on one side

Move everything to one side.

Each term changes its sign when it moves.

2 The characteristic equation

Powers are good at this: try aₙ = x to the n, for some number x.

Put x to the n in for aₙ, x to the n minus 1 for aₙ₋₁, and so on.

Divide everything by x to the n minus 2: the powers drop to 2, 1 and 0.

This is the characteristic equation.

Careful!

Watch the signs.

The terms moved to the left, so their signs changed.

3 Solve it

Factor: we need two numbers that multiply to −4 and add to −3: 1 and −4.

So x = −1 or x = 4.

4 The general form

Each root gives a power that works.

Any mix of them works too: A times −1 to the n, plus B times 4 to the n.

5 Use the start values

n = 0: every power to the 0 is 1.

So A + B = 2.

n = 1: −1A + 4B = 3.

From equation 1: A = 2 − B.

Put it into equation 2.

Open the bracket.

Like terms: 5B.

Take away −2 on both sides.

Divide by 5: B = 1.

And A = 2 minus B, that is 1.

So aₙ = (−1)ⁿ + 4ⁿ.

6 Check

Check a₂.

The rule: 3 times 3, plus 4 times 2: 17.

The formula with n = 2 gives 17 too.

Right!

Let's look back.

First: everything on one side.

Then: the characteristic equation.

Then: solve it.

Then: the general form.

Then: use the start values.

Last: check.