Proof by counterexample for $1+2+3+...+(n-1) = kn$

Question

I have a question that asks for a proof by counterexample that $1+2+3+...+(n-1) = kn$ for some $k\text{ } |k \in\Bbb Z $

This doesn't seem to be true for positive numbers, so I've used 4 as a counterexample.

when $n = 4$

$1 + 2 + 3 = k(4)$

$6 = 1.5(4)$

$k = 1.5 $

Since $k \notin \Bbb Z $

$1 + 2 + 3 + ... + (n-1) \neq kn \text{ for all } k | \in\Bbb Z $

Is this the correct way to do a proof by counterexample? How could I go about extending this to say that this equation doesn't hold for all positive values?

Answer 1

Your counterexample is correct. In general,

$1+2+3+...+(n-1) = \frac{n(n-1)}{2}=kn$,

Then $k=\frac{n-1}{2}$ which means that $k$ is an integer iff $n$ is odd.

Answer 2

Yes, $1+2+3+...+n-1=n(n+1)/2=kn$ so that $n$ is odd you can choose $k=(n-1)/2$