how to evaluate this series $\sum \limits_{k=1}^{n}\cosh(x+n.y) $?

Question

How to compute this series ? :

$\sum \limits_{k=1}^{n}\cosh (x+n.y) $ $~~~~~~~~~~~~~~~~~~~~/~~~~x,y \in\mathbb{R} ~~~~~~~~~~n\in \mathbb{N}$

where $\cosh$ is the hyperbolic cosine .

Answer 1

$$ \cosh(\theta) = \frac{\mathrm{e}^{\theta} +\mathrm{e}^{-\theta}}{2} $$ so your sum is $$ \sum_{k=0}^n\frac{\mathrm{e}^{x+ky} +\mathrm{e}^{-x-ky}}{2} $$ or $$ \mathrm{e}^{a+b} = \mathrm{e}^{a}\mathrm{e}^b\\ \mathrm{e}^{ab} =\left(\mathrm{e}^{b}\right)^a $$ so we have $$ \frac{1}{2}\mathrm{e}^x\sum_{k=0}^n\left(\mathrm{e}^{y}\right)^k+\frac{1}{2}\mathrm{e}^{-x}\sum_{k=0}^n\left(\mathrm{e}^{-y}\right)^k $$ a sum of the form $$ \sum_{k=0}^n a^k $$ is a geometric one.

Also note: $$ \sum_{k=0}^n a^k = 1 + \sum_{k=1}^n a^k \implies \sum_{k=1}^n a^k = -1 + \sum_{k=0}^n a^k $$