Tricky Summation

Calculus Level 3

For all integers nn, we define ξn\xi_n as follows: {ξn=1if n0(mod4) or n1(mod4)ξn=1if n2(mod4) or n3(mod4)\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 nZ+n \in \mathbb{Z^+}, let f(n)=ξ0(n0)+ξ1(n1)+ξ2(n2)++ξn(nn).f(n)= \xi_0 \dbinom{n}{0} + \xi_1 \dbinom{n}{1} + \xi_2 \dbinom{n}{2} + \cdots + \xi_n \dbinom{n}{n}. Find 100(n=0f(n)n!)\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 40(mod4)4 \equiv 0 \pmod{4}, ξ4=1\xi_4= 1, whereas ξ6=1\xi_6 = -1 since 62(mod4)6 \equiv 2 \pmod{4}. Note that ξ0=ξ1=1\xi_0= \xi_1= 1.

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

  • You might use a scientific calculator for this problem.

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