Show that the expression, $\frac{(n)(n+1)}{2}$ will never yield a multiple of $n$ for even values of $n$ where $n \neq 0$. $F(n) =\frac{(n)(n+1)}{2}$
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, $n = 4$
$F(4) =\frac{4(4+1)}{2}$
$F(4) = (2)(5)$

$n$ and $n+1$ are relatively prime, or perhaps more simply, if $n$ is even, then $n+1$ is odd.
