Steply

Solve a_n = 8a_(n-1) - 16a_(n-2), a_0 = 2, a_1 = 12

Step by step, with the reason for every step · Second-order recurrences · Discrete Math

Step by step

The question

Question: solve aₙ = 8aₙ₋₁ − 16aₙ₋₂, with a₀ = 2 and a₁ = 12.

an = 8an − 1 − 16an − 2, a0 = 2, a1 = 12

1 Everything on one side

Move everything to one side.

Each term changes its sign when it moves.

an − 8an − 1 + 16an − 2 = 0

2 The characteristic equation

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

try an = xn

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

xn − 8xn − 1 + 16xn − 2 = 0

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

This is the characteristic equation.

x2 − 8x + 16 = 0
÷ xn − 2 on both sides

Careful!

Watch the signs.

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

Careful: not x2 − 8x − 16 = 0 ✗

3 Solve it

Factor: we need two numbers that multiply to 16 and add to −8: −4 and −4.

(x − 4)² = 0
(−4) × (−4) = 16, (−4) + (−4) = −8

Both brackets are the same: one root, 4, twice.

x = 4 (twice)

4 The general form

With one root twice, the second part gets an extra n: A times 4 to the n, plus B times n times 4 to the n.

an = A × 4n + B × n × 4n

5 Use the start values

n = 0: the n in front makes the B part 0.

So A = 2.

A = 2 ①

n = 1: 4A + 4B = 12.

4A + 4B = 12 ②

Put A = 2 in.

8 + 4B = 12

Take away 8.

4B = 4

Divide by 4: B = 1.

B = 1

So aₙ = 2 × 4ⁿ + n × 4ⁿ.

an = 2 × 4n + n × 4n

6 Check

Check a₂.

The rule: 8 times 12, minus 16 times 2: 64.

rule: a2 = 8 × 12 − 16 × 2 = 64

The formula with n = 2 gives 64 too.

Right!

formula: a2 = 64

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.

Answer
an = 2 × 4n + n × 4n

Your turn

Try this one yourself, then check:

Solve an6an − 1 − 9an − 2, a0 = 3, a1 = 12
Show the answer
an = 3 × 3n + n × 3n

More from Second-order recurrences

The whole lesson: Second-order recurrences →