Show that $\sum_{n=0}^\infty(-1)^n \text{sech}{(F_n)}\text{sech}(F_{n+3})=\frac{1}{1+\cosh2}$

Question

Let $F_n$ be the $n^{th}$ Fibonacci number and $\text{sech}(x)=\frac{2}{e^{x}+e^{-x}}$.

How to show that $\sum_{n=0}^\infty(-1)^n \text{sech}{(F_n)}\text{sech}(F_{n+3})=\frac{1}{1+\cosh2}$?

Can we make it to be a telescopic sum ?

First I think each term can be write as $\frac{1}{\cosh(F_n)+\cosh(F_{n+3})}- \frac{1}{\cosh(F_{n+1})+\cosh(F_{n+4})}$ but it is not true.

Thanks in advance.