Main theme: We set for $n\geq 1$
\begin{align*}
A_n(x)=\sum_{s=0}^{n-1}\frac{Q_n(x)}{Q_s(x)}x^{s(2n+1)}\qquad\qquad
S_n(x)=\sum_{m=1}^{2n}x^{\frac{m(m-1)}{2}}\\
\end{align*}
and find a representation of $A_n(x)$ via $A_{n-1}(x)$ by also separating terms which are summands from $S_n(x)$.
\begin{align*}
\color{blue}{A_n(x)}&=A_{n-1}(x)+x^{(n-1)(2n-1)}+x^{n(2n-1)}\\
&\,\,\color{blue}{=A_{n-1}(x)+S_{n}(x)-S_{n-1}(x)}\tag{1}
\end{align*}
We so derive by induction
\begin{align*}
A_n(x)-S_n(x)&=A_{n-1}(x)-S_{n-1}(x)=\cdots\\
&=A_{1}(x)-S_{1}(x)\\
&=\frac{1-x^2}{1-x}-(1+x)=0
\end{align*}
and the claim follows.
Proof: The derivation of (1) is elementary but technically tricky.
We start with an algebraic identity. The following holds:
\begin{align*}
\left(1-x^{2n}\right)x^{s(2n+1)}&=\left(1-x^{2n-1}\right)x^{s(2n-1)}+\left(1-x^{2s+1}\right)x^{(s+1)(2n-1)}\tag{2}\\
&\qquad-\left(1-x^{2s}\right)x^{s(2n-1)}
\end{align*}
The identity (2) can be shown easily my multiplying out and comparing corresponding terms. The idea is to represent the left-hand side by an expression where $n$ is consequently replaced with $n-1$. This will be used as building block to obtain the sums in (1). Division of (2) by $1-x^{2n-1}$ gives
\begin{align*}
\frac{1-x^{2n}}{1-x^{2n-1}}x^{s(2n+1)}&=x^{s(2n-1)}+\frac{1-x^{2s+1}}{1-x^{2n-1}}x^{(s+1)(2n-1)}\tag{3}\\
&\qquad-\frac{1-x^{2s}}{1-x^{2n-1}}x^{s(2n-1)}
\end{align*}
Recalling $Q_n(x) = \prod_{r=1}^n \frac{1-x^{2r}}{1-x^{2r-1}}$ we have $$Q_n(x)=Q_{n-1}(x)\frac{1-x^{2n}}{1-x^{2n-1}}$$ We obtain from (3) by expanding with $\frac{Q_{n-1}(x)}{Q_s(x)}$:
\begin{align*}
\frac{Q_n{x)}}{Q_{s}(x)}x^{s(2n+1)}&=\frac{Q_{n-1}(x)}{Q_{s}(x)}x^{s(2n-1)}
+\underbrace{\frac{1-x^{2s+1}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}x^{(s+1)(2n-1)}}_{\alpha_{s,n}(x)}\\
&\qquad-\underbrace{\frac{1-x^{2s}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}}_{\beta_{s,n}(x)}\tag{4}
\end{align*}
Intermezzo:
We obtain for $s=0,\ldots,n-2$:
\begin{align*}
\color{blue}{\beta_{s+1,n}(x)}&=\frac{1-x^{2s+2}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_{s+1}(x)}x^{(s+1)(2n-1)}\\
&=\frac{1-x^{2s+2}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}\,\frac{1-x^{2s+1}}{1-x^{2s+2}}x^{(s+1)(2n-1)}\\
&=\frac{1-x^{2s+1}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_s(x)}x^{(s+1)(2n-1)}\\
&\,\,\color{blue}{=\alpha_{s,n}(x)}\tag{5}
\end{align*}
We also have $\beta_{0,n}(x)=0$ and
\begin{align*}
\alpha_{n-1,n}(x)=\frac{1-x^{2n-1}}{1-x^{2n-1}}\,\frac{Q_{n-1}(x)}{Q_{n-1}(x)}x^{n(2n-1)}=x^{n(2n-1)}\tag{6}
\end{align*}
Summing (4) from $s=0$ to $n-1$ we obtain with (5) and (6)
\begin{align*}
\color{blue}{A_n(x)}&=\sum_{s=0}^{n-1}\frac{Q_{n}(x)}{Q_s(x)}x^{s(2n+1)}\\
&=\sum_{s=0}^{n-1}\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}+\sum_{s=0}^{n-1}\alpha_{s,n}(x)-\sum_{s=1}^{n-1}\beta_{s,n}(x)\tag{7}\\
&=\sum_{s=0}^{n-1}\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}+\sum_{s=0}^{n-1}\alpha_{s,n}(x)-\sum_{s=0}^{n-2}\beta_{s+1,n}(x)\tag{8}\\
&=\sum_{s=0}^{n-1}\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}+\alpha_{n-1,n}(x)\tag{9}\\
&=\sum_{s=0}^{n-2}\frac{Q_{n-1}(x)}{Q_s(x)}x^{s(2n-1)}+x^{(n-1)(2n-1)}+x^{n(2n-1)}\tag{10}\\
&\,\,\color{blue}{=A_{n-1}(x)+S_{n}(x)-S_{n-1}(x)}
\end{align*}
which proves the claim. The other identity in (a) follows by adding the term with $s=n$ to both sides of the first identity.
Comment:
In (7) we use the identity (4) and sum from $s=0$ to $s=n-1$. We set the lower index of the right-most sum to $s=1$ since $\beta_{0,n}(x)=0$.
In (8) we shift the index of the right-most sum by one to start with $s=1$.
In (9) we use the telescoping property $\alpha_{s,n}(x)-\beta_{s+1,n}(x)=0$ for $s=0,\ldots, n-2$.
In (10) we separate the term with $s=n-1$ from the sum and use the identity (6).
Note: This proof was given by D. Shanks in Two Theorems of Gauss in 1958.
