what is this sum of squares of algebraic functions?

Question

This question is inspired by the MO query here, although it has no direct implications.

Define the family of polynomial functions $$f_n(x)=n^2x^{n-1}-\frac{d}{dx}\left(\frac{x^n-1}{x-1}\right),$$ and the associated family of algebraic functions $$g_n(x)=\frac{\sqrt{f_n(x)}}n+\sum_{j=1}^{n-1}\frac{\sqrt{f_j(x)}}{j(j+1)}.$$

Question. Despite the complicated expressions for $f_n$ and $g_n$, does $$h_n(x)=\sum_{k=0}^n\left(g_n(x)-g_k(x)\right)^2$$ have a neat (closed) formula?

UPDATE. user64494 (see below) has found $h_n(x)=nx^{n-1}$. Any proof?

Convention: $g_0:=0$ and empty sums are zero.

For example, $h_2(x)=g_2(x)^2+(g_2(x)-g_1(x))^2+(g_2(x)-g_2(x))^2=2x$.

Answer 1

A large part of the result doesn't really depend on $f_n$. For an arbitrary sequence $(a_n)$, define two sequences by $G_0=0$, $H_0=0$, and for $n\ge1$, $$G_n=\frac{a_n}{n}+\sum_{j=1}^n{\frac{a_j}{j(j+1)}}\quad\text{and}\quad H_n=\sum_{k=0}^n{(G_n-G_k)^2}.$$ Then $$H_{n+1}-H_n=G_{n+1}^2-G_n^2+\sum_{k=1}^n{(G_{n+1}-G_n)(G_{n+1}+G_n-2G_k)}.$$ The right-hand side simplifies to $$(G_{n+1}-G_n)\big((n+1)(G_{n+1}+G_n)-2\sum_{k=1}^n{G_k}\big),$$ where the last sum simplifies a lot: $$\sum_{k=1}^n{G_k}=(n+1)\sum_{j=1}^n{\frac{a_j}{j(j+1)}}$$ so that in the end everything telescopes and $$H_{n+1}-H_n=\frac{1}{n+1}(a_{n+1}^2-a_n^2).$$ Applying this with $a_n=\sqrt{f_n(x)}$ gives $$h_{n+1}(x)-h_n(x)=\frac{1}{n+1}(f_{n+1}(x)-f_n(x))=\frac{1}{n+1}\left((n+1)^2x^n-n^2x^{n-1}-\frac{d}{dx}\left(\frac{x^{n+1}-x^n}{x-1}\right)\right)= \frac{1}{n+1}\left((n+1)^2x^n-n(n+1)x^{n-1}\right)=(n+1)x^n-nx^{n-1},$$ whence the proof by induction.

Answer 2

The results of the Mathematica's code

f := Function[{n, x}, n^2*x^(n - 1) - D[(x^n - 1)/(x - 1), x]]
g := Function[{n, x},Sqrt[f[n, x]]/n + Sum[Sqrt[f[j, x]]/(j*(j + 1)), {j, 1, n - 1}]]
n = 60; FullSimplify[ g[n, x]^2 + Sum[(g[n, x] - g[k, x])^2, {k, 1, n}]]

60 x^59

n = 34; FullSimplify[g[n, x]^2 + Sum[(g[n, x] - g[k, x])^2, {k, 1, n}]]

34 x^33

suggest the answer $nx^{n-1}$. The same result is obtained with Maple 2016 through

restart; f := (n, x)-> n^2*x^(n-1)-(diff((x^n-1)/(x-1), x)):
g := (n, x) -> sqrt(f(n, x))/n+sum(sqrt(f(j, x))/(j*(j+1)), j = 1 .. n-1) :
simplify(expand([seq(sum((g(N, x)-g(k, x))^2, k = 1 .. N)+g(N, x)^2, N = 1 .. 5)]));

[1, 2*x, 3*x^2, 4*x^3, 5*x^4]