# Prove that the product of 2 consecutive natural numbers is even

1. Dec 3, 2012

### Accretion

1. The problem statement, all variables and given/known data
Prove that the product of two consecutive natural numbers is even.

2. The attempt at a solution
Hi, I'm just starting to work with proofs by induction, I'm just wondering if this is a valid technique, and/or if I am being too verbose in my proof, thanks!

Let $P(n)=n(n+1)$ is even
Let $f(n)=n(n+1)$

Let $n=1$
Since $1(1+1)=2$, and 2 is even, it follows that $P(1)$ is true.
Suppose $P(k)$ is true, that is, suppose $k(k+1)$ is even.
Then,
$f(k+1)-f(k)=(k+1)((k+1)+1)-k(k+1)$
$f(k+1)-f(k)=(k+1)(k+2)-k(k+1)$
$f(k+1)-f(k)=(k+1)(k+2-k)$
$f(k+1)-f(k)=2(k+1)$
$f(k+1)=2(k+1)+f(k)$

Since by our induction hypothesis, $f(k)$ is even, it follows that the sum, $f(k+1)$, of $2(k+1)$ and $f(k)$ is also even. Since $P(1)$ is true, and $P(k+1)$ is true if $P(k)$ is true, by induction, $P(n)$ is true for all natural numbers $n$.

Last edited: Dec 3, 2012
2. Dec 3, 2012

### lurflurf

That is right. You might not need so many algebra steps. The telescoping sum is just a fancy way to see that

$$n(n+1)=2 \sum_{k=1}^n k$$

3. Dec 3, 2012

### Mentallic

If you're going to be explaining P(k) is even by assumption, why not explicitly make it so? Let P(k) = k(k+1) = 2n where n is some natural number, so then you can plug that in later to obtain

P(k+1) = 2(k+1+n)

And then say, since n and k are assumed to be natural numbers, P(k+1) is even... etc.