Distance from point to subset of real binominals Prove that in $$\mathbb{R}_n[x]$$ with dot product defined as $$\langle P,Q \rangle =\int_0^1 P(x)Q(x) \, dx$$ distance from $$x^n$$ to $$\operatorname{span}(1,x,\ldots,x^{n-1})$$ is equal to $$\left({2n \choose n}\sqrt{2n-1} \right)^{(-1)}$$
I tried to solve the task with Gram matrix and distance between point and set defined by them, but I stuck when it got to counting the determinant of $(n+1)\times(n+1)$ matrix. Also I tried induction, or finding some kind of recursion, but... again got nowhere. I will appreciate any hint or help to solve that problem. 
 A: The correct result is $\left({2n \choose n}\sqrt{2n+1} \right)^{-1}$.
There is a vector $a=(a_j)_{0\leq j\leq n-1}$ s.t.  for every $0\leq i\leq n-1$,  $x^n-\sum_{j=0}^{n-1}a_jx^j$ is $\perp$ to $x^i$, that is $\dfrac{1}{n+i+1}=\sum_{j=0}^{n-1}a_j\dfrac{1}{i+j+1}$ or, equivalently, $H_na=u$ where $H_n$ is the Hilbert matrix of dimension $n$ and $u=[\dfrac{1}{n+1},\cdots,\dfrac{1}{2n}]^T$.
Let $y=[1,x,\cdots,x^{n-1}]^T$. The required distance $d_n$ satisfies $d_n^2=\langle x^n-a^Ty,x^n-\sum_ja_jx^j \rangle=\langle x^n-a^Ty,x^n \rangle$, that is $d_n^2=\dfrac{1}{2n+1}-\sum_{j=0}^{n-1}a_j\dfrac{1}{j+n+1}=\dfrac{1}{2n+1}-a^Tu=\dfrac{1}{2n+1}-u^TH_n^{-1}u$.
The sequence $z_n=d_n^{-2}$ has the number A000515 in the Sloane on-line encyclopedia and its value is $(2n+1)\binom{2n}{n}^2$.
cf. https://oeis.org/search?q=A000515+&sort=&language=english&go=Search
EDIT. We can also reason as follows. $d_n^2=\dfrac{Gram(1,x,\cdots,x^n)}{Gram(1,x,\cdots,x^{n-1})}=\dfrac{\det(H_{n+1})}{\det(H_n)}$. Note that $d_n^{-2}=\dfrac{\det(H_{n})}{\det(H_{n+1})}$ is the $(n+1,n+1)$ entry of $H_{n+1}^{-1}$. The values of the $1/\det(H_n)$ are given here 
https://oeis.org/A005249
