Prove 1 + 3 + 5 + ... + (2n - 1) = n^2 by induction Step by Step

Step by step

The question

Question: prove by induction that 1 + 3 + 5 + … + (2n − 1) = n².

1 + 3 + 5 + … + (2n − 1) = n²

Induction works like dominoes.

Push the first one: it is true for n = 1.

Then show each one knocks over the next: if it is true for k, it is true for k + 1.

Then they all fall.

1 Check n = 1

For n = 1 the left side is just the first number, 1.

The right side with n = 1 is also 1.

The first domino falls.

1 = 1² = 1

✓ n = 1 works

2 Assume it works for k

Now suppose it works for some whole number k: the sum up to k equals the formula with k.

1 + 3 + 5 + … + (2k − 1) = k²

we assume this

3 What we must show

We must show it for k + 1: one more term, (2k + 1), and every n on the right becomes k + 1.

1 + 3 + 5 + … + (2k − 1) + (2k + 1) = (k + 1)²

we must get this

Careful!

k + 1 goes into every n of the formula.

Adding 1 at the end is not the same.

Careful: not k² + 1 ✗

4 Use the assumption

The left side starts with the sum up to k.

By our assumption, that is the formula with k.

We swap it in.

1 + 3 + 5 + … + (2k − 1) + (2k + 1) = k² + (2k + 1)

5 Make both sides the same

Open the brackets.

= k² + 2k + 1

Now the right side, the one we must get.

The square means the bracket times itself.

(k + 1)² = (k + 1)(k + 1)

Open the brackets.

= k² + k + k + 1

Like terms together.

= k² + 2k + 1

The same as the left side!

So if it works for k, it works for k + 1.

= the left side ✓

It is true for n = 1, and each n passes it on to the next.

So it is true for every whole number n.

The proof is done.

true for n = 1, and k → k + 1: true for every n ≥ 1. ∎

6 Check

Let's check n = 4 by adding: 1 + 3 + 5 + 7 is 16.

n = 4: 1 + 3 + 5 + 7 = 16

And the formula with 4: also 16.

Right!

4² = 16

Let's look back.

First: check n = 1.

Then: assume it works for k.

Then: what we must show.

Then: use the assumption.

Then: make both sides the same.

Last: check.

Answer