$F(n) =\frac{(n)(n+1)}{2}$
Show that $F(n)$ will never yield a multiple of $n$ for even values of $n$ where $n \neq 0$.
Let, $n = 2$
$F(2) =\frac{2(2+1)}{2}$
$F(2) = 3$
Let, $n = 4$
$F(4) =\frac{4(4+1)}{2}$
$F(4) = (2)(5)$
$F(n) =\frac{(n)(n+1)}{2}$
Show that $F(n)$ will never yield a multiple of $n$ for even values of $n$ where $n \neq 0$.
Let, $n = 2$
$F(2) =\frac{2(2+1)}{2}$
$F(2) = 3$
Let, $n = 4$
$F(4) =\frac{4(4+1)}{2}$
$F(4) = (2)(5)$
One can say it's obvious, and it is, sort of. But if it's obvious we should be able to prove it!
Suppose to the contrary that $\dfrac{n(n+1)}{2}$ is a multiple of $n$. Then $\frac{n(n+1)}{2}=nk$ for some integer $k$. Equivalently, $n(n+1)=2nk.$ Now it is tempting to cancel $n$, which we can, unless $n=0$.
Note that in fact if $n=0$, then $\dfrac{n(n+1)}{2}$ is a multiple of $n$. So we have found that the assertion is not quite true. There is an even value of $n$, namely $n=0$, at which things break down.
Suppose now that $n\ne 0$. Then we can divide both sides by $n$, obtaining $2k=n+1.$ This is impossible, the left side is even, and $n+1$ is odd.
$n$ and $n+1$ are relatively prime, or perhaps more simply, if $n$ is even, then $n+1$ is odd.