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

Step by step

The question

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

a_n = 2a_{n − 1} + 1, a₁ = 1

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 = 1.

start

Then 2 times 1, plus 1.

2 × 1 + 1

That is 3.

And so on.

2 × 3 + 1

2 × 7 + 1

2 × 15 + 1

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 2c + 1, and it must be c again.

c = 2c + 1

Careful!

c is on both sides.

We solve for it like any equation.

Careful: not c = 2 + 1 = 3 ✗

Take away 2c on both sides.

c − 2c = 1

c minus 2c is −c.

−c = 1

Divide by −1: c = −1.

c = −1

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

check: 2 × (−1) + 1 = −1 ✓

3 The distance from it

Now the distance from c: each term plus 1.

Look at the new column: 2, 4, 8, 16, 32.

Each one is 2 times the one before.

The distance from c grows by times 2.

× 2 each time

4 The formula

The distance starts at 2 and is multiplied by 2 each step: 2 times 2 to the n − 1.

a_n + 1 = 2 × 2^{n − 1}

Take away 1 on both sides.

a_n = 2 × 2^{n − 1} − 1

2 is one more 2, so the powers join: 2 times 2 to the n − 1 is 2 to the n.

= 2ⁿ − 1

5 Check

Check n = 4 with the formula: it gives 15.

n = 4: formula gives 15

The table says 15 too.

Right!

table: a₄ = 15

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