Step by step
The question
Question: prove by induction that 2n³ + 3n² + n is divisible by 6, for every whole number n.
Like dominoes: show it for n = 1, then show that k passes it on to k + 1.
1 Check n = 1
For n = 1: 6, and that is 6 × 1.
Divisible by 6: the first domino falls.
2 Assume it works for k
Suppose it works for k: 2k³ + 3k² + k is a multiple of 6.
So it is 6 times some whole number.
Call it m.
3 What we must show
We must show the same for k + 1: every n becomes k + 1.
Careful!
A bracket to a power is the bracket times itself.
Every part times every part.
4 Use the assumption
Each power of the bracket is the bracket times itself.
Open the brackets.
Like terms together.
Look for the old one inside: 2k³ + 3k² + k.
The rest is left over.
By our assumption, the old one is 6m.
5 Take out 6
Every term has a 6 in it.
Take out 6: 6(k² + 2k + m + 1).
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.
6 Check
Let's check n = 3: it gives 84.
84 is 6 times 14.
Divisible by 6.
Right!
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 6.
Last: check.