Proof involving Euler's Totient Function

Question

Let n $\ge$ 2 be a natural number, k and n are coprime. Prove

$\sum_{1\le k\le n}k$ = $\frac{n\phi (n)}{2}$

Answer 1

As is pointed out in the comments $(k,n)=(n-k,n)$ so for every $k$ in your sum you'll also have $n-k$. So $$\sum_{1\leq k\leq n}k=\sum_{1\leq k\leq \frac{n}{2}}(k+n-k)=\frac{\phi(n)n}{2}$$

Notice that we cannot have $k=n-k$, since $n=2k$ would imply that they aren't coprime.