Express $C_n = \cosh(0) + \cosh(1) + \cosh (2) + \dots + \cosh(n)$ Could someone give me some hint of how to do this question please.

I've been stuck for more than $3$ hours on this question.
 A: Replacing $x$ with $k$ in the given relations (and rearranging somewhat), we have
$$\begin{cases}
&\sinh{(1)}\cosh{(k)}+\left(\cosh{(1)}-1\right)\sinh{(k)}=\sinh{(k+1)}-\sinh{(k)},\\
&\left(\cosh{(1)}-1\right)\cosh{(k)}+\sinh{(1)}\sinh{(k)}=\cosh{(k+1)}-\cosh{(k)}.\\
\end{cases}$$
Summing both equations from $k=0$ to $k=n$, we find
$$\begin{cases}
&\sinh{(1)}C_{n}+\left(\cosh{(1)}-1\right)S_{n}=\sinh{(n+1)},\\
&\left(\cosh{(1)}-1\right)C_{n}+\sinh{(1)}S_{n}=\cosh{(n+1)}-1.\\
\end{cases}$$
This yields a system of two linear equations in two unknowns, $C_{n}$ and $S_{n}$, which can be solved by standard methods. The result is
$$\begin{cases}
&C_{n}=\frac{\sinh{(1)}\sinh{(n+1)}-\left(\cosh{(1)}-1\right)\left(\cosh{(n+1)}-1\right)}{\sinh^2{(1)}-\left(\cosh{(1)}-1\right)^2},\\
&S_{n}=\frac{\sinh{(1)}\left(\cosh{(n+1)}-1\right)-\left(\cosh{(1)}-1\right)\sinh{(n+1)}}{\sinh^2{(1)}-\left(\cosh{(1)}-1\right)^2}.\\
\end{cases}$$
Using the hyperbolic Pythagorean identity, $\sinh^2{(x)}-\cosh^2{(x)}=-1$, the denominators can be simplified as
$$\begin{align}
\sinh^2{(1)}-\left(\cosh{(1)}-1\right)^2
&=\sinh^2{(1)}-\left(\cosh^2{(1)}-2\cosh{(1)}+1\right)\\
&=\sinh^2{(1)}-\cosh^2{(1)}+2\cosh{(1)}-1\\
&=2\cosh{(1)}-2\\
&=2\left(\cosh{(1)}-1\right).\\
\end{align}$$
If one desires to eliminate $\sinh{(1)}$ from the numerator as well, one can simply replace it with $\sinh{(1)}=\sqrt{\cosh^2{(1)}-1}$, again using the Pythagorean relation.
