s$.

\[lema03\] $$\sum_{k=2}^{s} \binom{r}{k} (k-1) \left(\frac{1+(-1)^{k}}{2}\right)
=1+2^{s-2}(s-2).$$

Consider the function $$f(x)=\frac{(1+x)^{s}}{2x}=\frac{1}{2}\sum_{k=0}^{s} \binom{s}{k} x^{k-1}$$ and its derivative $$f'(x)=\frac{(1+x)^{s-1}((s-1)x-1)}{2x^{2}}=\frac{1}{2}\sum_{k=0}^{s} \binom{s}{k} (k-1) x^{k-2}.$$

Next, we evaluate at $x=1$ and $x=-1$ and then we add the results to obtain $$2^{s-2}(s-2)=f'(1)+f'(-1)=\frac{1}{2}\sum_{k=0}^{s} \binom{s}{k} (k-1) (1+(-1)^{k}) =-1+\sum_{k=2}^{s} \binom{s}{k} \frac{(1+(-1)^{k})}{2} (k-1).$$

We may apply Proposition \[coroKR\] for $C$ using all the subgroups $H_{i_{1},\ldots,i_{k}}$ in order to obtain that $JC$ (so $JS$) is isogenous to the product of the jacobian varieties of all Riemann surfaces $C/H_{i_{1},\ldots,i_{k}}$ as desired. The equations of these curves are provided in Lemma \[lema02\].

Proof of Theorem \[coro3\]
==========================

As a consequence of Theorem \[construccion\] (and Lemma \[lema02\]), the jacobian variety of $S$ is isogenous to the product of the following six elliptic curves $$\begin{array}{ll}
E_{1}: & y^{2}=x(x-1)(x-\lambda)\\
E_{2}: & y^{2}=(x-1)(x-\lambda)(x-\mu_{1,1})(x-\mu_{1,2})\\
E_{3}: & y^{2}=(x-\mu_{1,1})(x-\mu_{1,2})(x-\mu_{2,1})(x-\mu_{2,2})\\
E_{4}: & y^{2}=x(x-\mu_{2,1})(x-\mu_{2,2})\\
E_{5}: & y^{2}=x(x-\mu_{1,1})(x-\mu_{1,2})\\
E_{6}: & y^{2}=(x-1)(x-\lambda)(x-\mu_{2,1})(x-\mu_{2,2})
\end{array}$$

and the jacobian variety of the following genus three hyperelliptic Riemann surface $$R: y^{2}=x(x-1)(x-\lambda)(x-\mu_{1,1})(x-\mu_{1,2})(x-\mu_{2,1})(x-\mu_{2,2}).$$

Let us observe that the group $J=\langle f_{1}(x)=\lambda/x, f_{2}(x)=\lambda (x-1)/(x-\lambda) \rangle \cong {\mathbb Z}_{2}^{2}$ keeps invariant the collection $$\infty, 0, 1, \lambda, \mu_{1,1}, \mu_{1,2}, \mu_{2,1}, \mu_{2,2}.$$

In this way, $R$ admits the following automorphisms $$F_{1}(x,y)=\left( \frac{\lambda}{x}, \frac{\lambda^{2} y}{x^{4}}\right)$$ $$F_{2}(x,y)=\left( \frac{\lambda(x-1)}{x-\lambda}, \frac{\lambda^{2}(\lambda-1)^{2} y}{(x-\lambda)^{4}}\right)$$

We may see that $\langle 