# Tricky Summation

**Calculus**Level 5

For all integers \(n\), we define \(\xi_n\) as follows: \[\begin{cases} \xi_n = 1 & \text{if } n \equiv 0 \pmod{4} \text{ or } n \equiv 1 \pmod{4} \\ \xi_n= -1 & \text{if } n \equiv 2 \pmod{4} \text{ or } n \equiv 3 \pmod{4} \end{cases} \] For all \(n \in \mathbb{Z^+}\), let \[f(n)= \xi_0 \dbinom{n}{0} + \xi_1 \dbinom{n}{1} + \xi_2 \dbinom{n}{2} + \cdots + \xi_n \dbinom{n}{n}.\] Find \(\left \lfloor 100 \left( \displaystyle \sum \limits_{n=0}^{\infty} \dfrac{f(n)}{n!} \right) \right \rfloor\).

**Details and assumptions**

As an explicit example, since \(4 \equiv 0 \pmod{4}\), \(\xi_4= 1\), whereas \(\xi_6 = -1\) since \(6 \equiv 2 \pmod{4}\). Note that \(\xi_0= \xi_1= 1\).

The floor function \(\lfloor x \rfloor \) denotes the largest integer less than or equal to \(x\). For example, \(\lfloor 3.25 \rfloor = 3, \lfloor 4 \rfloor= 4, \lfloor \pi \rfloor = 3\).

You might use a scientific calculator for this problem.