Prove 7^n - 2^n is divisible by 5 by induction Step by Step

Step by step

The question

Question: prove by induction that 7ⁿ − 2ⁿ is divisible by 5, for every whole number n.

7ⁿ − 2ⁿ is divisible by 5

Like dominoes: show it for n = 1, then show that k passes it on to k + 1.

1 Check n = 1

For n = 1: 5, and that is 5 × 1.

Divisible by 5: the first domino falls.

7¹ − 2¹ = 5 = 5 × 1

2 Assume it works for k

Suppose it works for k: 7ᵏ − 2ᵏ is a multiple of 5.

So it is 5 times some whole number.

Call it m.

7^k − 2^k = 5m

m is a whole number

Move the rest to the other side: 7ᵏ equals 2ᵏ + 5m.

We will need this.

7^k = 2^k + 5m

3 What we must show

We must show the same for k + 1: every n becomes k + 1.

= 7^{k + 1} − 2^{k + 1} is divisible by 5

Careful!

One more power means one more times the number.

It is 7 times as big, not 1 more.

Careful: not 7^{k + 1} = 7^k + 1 ✗

4 Use the assumption

One more power: 7 times the power with k.

7^{k + 1} − 2^{k + 1} = 7 × 7^k − 2 × 2^k

7^{k + 1} = 7 × 7^k

Now swap in 2ᵏ + 5m for 7ᵏ.

= 7(2^k + 5m) − 2 × 2^k

Open the bracket.

= 35m + 7 × 2^k − 2 × 2^k

Like terms together.

= 5 × 2^k + 35m

5 Take out 5

Every term has a 5 in it.

Take out 5: 5(2ᵏ + 7m).

= 5(2^k + 7m)

2ᵏ + 7m is a whole number

So it works for k + 1 too.

True for n = 1, and passed on each time: true for every whole number n.

The proof is done.

divisible by 5 for n = 1, and k → k + 1: for every n. ∎

6 Check

Let's check n = 3: it gives 335.

n = 3: 7³ − 2³ = 335

335 is 5 times 67.

Divisible by 5.

Right!

335 = 5 × 67

Let's look back.

First: check n = 1.

Then: assume it works for k.

Then: what we must show.

Then: use the assumption.

Then: take out 5.

Last: check.

Answer