# Prove product of 3 consecutive naturals is even

1. Dec 4, 2012

### Mouse07

1. Prove product of 3 consecutive naturals is even data

i just want to know if this proof is acceptable, or is there a simpler proof.

3. The attempt at a solution

Proof: suppose if the product 3 consecutive naturals is dividable by 2 then it is even.

Base case: n = 1

n (n + 1) (n + 2 )
1(2)(3)
6
since 6 is dividable by 2 base case holds

Inductive Hypo If n holds then (n+1)

inductive step

(n + 1) (n + 2)(n+3)
(n^2+3n+2)(N+3)
n^3 + 6n^2 + 11n + 6

n(n^2 + 11)+ 6 (n^2 +1)

since n (n^2 + 11) is even and 6 (n^2 + 1) is even,

and even + even = even.

since all even numbers are divisible by 2, thus n +1 hold.

hence the product of 3 consecutive naturals is even.
QED(i know that some sentences are missing but that's how i did the proof)

2. Dec 4, 2012

### Staff: Mentor

Where is your induction hypothesis? You will need it in the inductive step.
Why is n(n2 + 11) even? You can't just say this.
There's a much simpler way to prove that n(n + 1)(n + 2) is even. All you need are two cases:
1. Assume that n is even.
2. Assume that n is odd.

3. Dec 4, 2012

### Mouse07

Ohh lol i thought about it, but it felt to simple so didn't do that.

Why is n(n2 + 11) even?

Because if n is odd then n^2 + 11 is even and even multiple by odd is even.
if n is even then even multiple by odd is even. I should add that but, now just gone do what you said it easier.

4. Dec 4, 2012

### Mentallic

Then explicitly show this.
If n is odd, thus n = 2k+1 then

$$n(n^2+11) = (2k+1)((2k+1)^2+11)$$

$$=(2k+1)(4k^2+4k+1+11)$$

$$=(2k+1)(4k^2+4k+12)$$

$$=2(2k+1)(2k^2+2k+6)$$

Which is even... etc.

5. Dec 5, 2012

### dextercioby

The simplest proof would require the solver to see that a sufficient condition for the product of 3 terms to be even is that the simplest subproduct be even, i.e. n(n+1).