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

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

Step by step

The question

Question: prove by induction that n³ + 2n is divisible by 3, for every whole number n.

n3 + 2n is divisible by 3

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

1 Check n = 1

For n = 1: 3, and that is 3 × 1.

Divisible by 3: the first domino falls.

13 + 2 × 1 = 3 = 3 × 1

2 Assume it works for k

Suppose it works for k: k³ + 2k is a multiple of 3.

So it is 3 times some whole number.

Call it m.

k3 + 2k = 3m
m is a whole number

3 What we must show

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

= (k + 1)3 + 2(k + 1) is divisible by 3

Careful!

A bracket to a power is the bracket times itself.

Every part times every part.

Careful: not (k + 1)³ = k³ + 1 ✗

4 Use the assumption

Each power of the bracket is the bracket times itself.

(k + 1)3 + 2(k + 1) = (k + 1)(k + 1)(k + 1) + 2(k + 1)

Open the brackets.

= k3 + k2 + k2 + k + k2 + k + k + 1 + 2k + 2

Like terms together.

= k3 + 3k2 + 5k + 3

Look for the old one inside: k³ + 2k.

The rest is left over.

= (k3 + 2k) + 3k2 + 3k + 3

By our assumption, the old one is 3m.

= 3m + 3k2 + 3k + 3

5 Take out 3

Every term has a 3 in it.

Take out 3: 3(k² + k + m + 1).

= 3(k2 + k + m + 1)
k² + k + m + 1 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 3 for n = 1, and k → k + 1: for every n. ∎

6 Check

Let's check n = 3: it gives 33.

n = 3: 33 + 2 × 3 = 33

33 is 3 times 11.

Divisible by 3.

Right!

33 = 3 × 11

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 3.

Last: check.

Answer
k² + k + m + 1 is a whole number

Your turn

Try this one yourself, then check:

2n3 + 3n2 + n is divisible by 6
Show the answer
6(k2 + 2k + m + 1)

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