Combinatorial Summation I'm trying to solve the following question:

If $s_n$ is the sum of first $n$ natural numbers, then prove that
$$2(s_1s_{2n}+s_2s_{2n-1}+\dots+s_ns_{n+1})=\frac{(2n+4)!}{5!(2n-1)!}$$
This is where I've come so far:
The general term of $(1-x)^{-2n}$ happens to be $t_{r+1}=\frac{(2n+r-1)!}{(2n-1)!r!}x^r$ and therefore the 6th term, i.e, $t_{5+1}=t_6=\frac{(2n+4)!}{(2n-1)!5!}x^5$, which looks like the RHS of the question I'm trying to solve.
Second, I found that $$(1-x)^{-3}=s_1+s_2x+\dots+s_nx^{n-1}+\dots$$
It's clear that the LHS must be something like, $(1-x)^{-a}(1-x)^{-b}$. But I'm not able to find one such combination which would give me $2(s_1s_{2n}+s_2s_{2n-1}+\dots+s_ns_{n+1})$.
Help is greatly appreciated.

If you'd be open to another strategy, we want to prove$$\sum_{k=1}^ns_ks_{2n+1-k}=\frac{n(n+1)(n+2)(2n+1)(2n+3)}{30}.$$This is an exercise in arithmetic once you notice$$s_k=\tfrac12k(k+1)\implies s_ks_{2n+1-k}=\tfrac14k(k+1)(2n+1-k)(2n+2-k)$$is a quartic in $k$, and use the Faulhaber's formulae$$\begin{align}\sum_{k=1}^nk&=\tfrac12n(n+1),\\\sum_{k=1}^nk^2&=\tfrac16n(n+1)(2n+1),\\\sum_{k=1}^nk^3&=\tfrac14n^2(n+1)^2,\\\sum_{k=1}^nk^4&=\tfrac{1}{30}n(6n^4+15n^3+10n^2-1).\end{align}$$
