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Prove 3^n + 7 is divisible by 2 by induction

Step by step, with the reason for every step · Induction · Discrete Math

Step by step

The question

Question: prove by induction that 3ⁿ + 7 is divisible by 2, for every whole number n.

3n + 7 is divisible by 2

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

1 Check n = 1

For n = 1: 10, and that is 2 × 5.

Divisible by 2: the first domino falls.

31 + 7 = 10 = 2 × 5

2 Assume it works for k

Suppose it works for k: 3ᵏ + 7 is a multiple of 2.

So it is 2 times some whole number.

Call it m.

3k + 7 = 2m
m is a whole number

Move the rest to the other side: 3ᵏ equals 2m − 7.

We will need this.

3k = 2m − 7

3 What we must show

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

= 3k + 1 + 7 is divisible by 2

Careful!

One more power means one more times the number.

It is 3 times as big, not 1 more.

Careful: not 3k + 1 = 3k + 1 ✗

4 Use the assumption

One more power: 3 times the power with k.

3k + 1 + 7 = 3 × 3k + 7
3k + 1 = 3 × 3k

Now swap in 2m − 7 for 3ᵏ.

= 3(2m − 7) + 7

Open the bracket.

= 6m − 21 + 7

Like terms together.

= 6m − 14

5 Take out 2

Every term has a 2 in it.

Take out 2: 2(3m − 7).

= 2(3m − 7)
3m − 7 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 2 for n = 1, and k → k + 1: for every n. ∎

6 Check

Let's check n = 3: it gives 34.

n = 3: 33 + 7 = 34

34 is 2 times 17.

Divisible by 2.

Right!

34 = 2 × 17

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 2.

Last: check.

Answer
3m − 7 is a whole number

Your turn

Try this one yourself, then check:

7n − 2n is divisible by 5
Show the answer
5(2k + 7m)

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