Fibonacci Complex Sum
The Fibonacci sequence is defined by $ F_1 = F_2 = 1 $ and $ F_n = F_{n - 1} + F_{n - 2} $ for $ n \ge 3 $.
Compute
\[\sum_{j = 1}^{2004} i^{2004 - F_j}.\]
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Since $ i^4 = 1, $ $ i^{2004 - F_j} = \frac{1}{i^{F_j}} $ depends only on the value of $ F_j $ modulo 4.
We compute the first few Fibonacci numbers modulo 4:
\[
\begin{array}{c|c}
n & F_n \pmod{4} \\ \hline
1 & 1 \\
2 & 1 \\
3 & 2 \\
4 & 3 \\
5 & 1 \\
6 & 0 \\
7 & 1 \\
8 & 1
\end{array}
\]Since $ F_7 \equiv F_1 \equiv 1 \pmod{4} $ and $ F_8 \equiv F_2 \equiv 1 \pmod{4}, $ and each term depends only on the previous two terms, the Fibonacci numbers modulo 4 becomes periodic, with period 6.
Since $ 2004 = 334 \cdot 6, $
\[\sum_{j = 1}^{2004} \frac{1}{i^{F_j}} = 334 \left( \frac{1}{i} + \frac{1}{i} + \frac{1}{i^2} + \frac{1}{i^3} + \frac{1}{i} + \frac{1}{1} \right) = \boxed{-668i}.\]
Freeflows