To all the genius out there , here is a question about expresssing summation of hyperboilc functions :
First of all, I've already proved that:
$$\sinh(x + 1)- \sinh(x) = (-1 + \cosh (1)) \sinh(x) + \sinh (1) \cosh(x)$$
$$\cosh(x + 1)- \cosh(x) = (-1 + \cosh (1)) \cosh(x) + \sinh (1) \sinh(x)$$
The later question asked me to use these results and express the following sums:
$C_n = \cosh (0) + \cosh (1) + \cosh (2) + \ldots +\cosh (n)$
$S_n = \sinh (0) + \sinh (1) + \sinh (2) + \ldots+\sinh (n)$
I have learned a few summation of hyberbolic function by myself, please guide me through this new challenge , thank you .
For simplicity of derivation, let $c = \cosh 1$ and $s = \sinh 1$, we have
$$\begin{align} \sinh(x+1) - \sinh x &= (c - 1) \sinh x + s \cosh x \tag{*1a}\\ \cosh(x+1) - \cosh x &= (c - 1) \cosh x + s \sinh x \tag{*1b} \end{align}$$ $s \times (*1a) - (c-1) \times (*1b) \implies$ $$ \begin{align} & s \text{LHS}(*1a) - (c-1)\text{LHS}(*1b)\\ =\; & s\left( (c - 1) \sinh x + s \cosh x \right) -(c-1)\left( (c - 1) \cosh x + s \sinh x \right)\\ =\; & (s^2 -(c-1)^2)\cosh x\\ =\; & 2(c-1)\cosh x \end{align} $$ This leads to $$\cosh x = \frac12 \left(\frac{s}{c-1}\text{LHS}(*1a) - \text{LHS}(*1b)\right) = A(x+1) - A(x)$$ where $$A(x) = \frac12 \left(\frac{s}{c-1}\sinh x - \cosh x\right)$$ From this, we can conclude
$$\begin{align} C_n = \sum_{k=0}^n \cosh k &= \sum_{k=0}^n ( A(k+1) - A(k) ) = A(n+1) - A(0)\\ &= \frac12 \left(\frac{\sinh(1)}{\cosh(1)-1}\sinh(n+1) - (\cosh(n+1) - 1 )\right) \end{align} $$
The derivation of $S_n$ is similar, I will skip it here.