1
$\begingroup$

$F(n) =\frac{(n)(n+1)}{2}$

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, $n = 4$

$F(4) =\frac{4(4+1)}{2}$

$F(4) = (2)(5)$

  • 1
    This one is straight forward. What have you tried? Note that you require $n > 0$ here.2012-12-12
  • 5
    35 questions, still disregarding completely the proper way to ask them.2012-12-12

2 Answers 2

7

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.

0

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