probability - If $mathbb P[X_n=k]=C_nmax(k,n-k)$ for $k=1,...n-1$, then find $C_n$

For an arbitrary $n\ge 3$, let $X_n$ be a random variable on $\{1,2,...,n-1\}$, whose distribution is

$\mathbb P[X_n=k]=C_n\max(k,n-k)$ for $k=1,...n-1$

What is the value of constant $C_n$ ?

I tried it, but obtained a strange result

So the necessary condition is $\displaystyle\sum\limits_{k=1}^{n-1}\mathbb P[X_n=k]=1$

I tried it for small numbers and i noticed that the summands are symmetric, It doesn't matter if $n$ is odd or even, the formula i got is for the sum without $C_n$

$\Big(2\displaystyle\sum\limits_{i=1}^{\lceil\frac{n-1}{2}\rceil}n-i\Big)-\Big(\frac{n-1}{2}-\lfloor{\frac{n-1}{2}}\rfloor\Big)n$

If $n$ is odd then the term after the sum vanishes.

And $C_n$ is then the reciprocal of this expression above.

Maybe we can simplify this ?