Řešení teoretické úlohy 6

Nechť $φ = \frac{1 + \sqrt{5}}{2}, ϕ = \frac{1 - \sqrt{5}}{2}$ . Platí:

φ^{2} = {(\frac{1 + \sqrt{5}}{2})}^{2} = \frac{1 + 2 \sqrt{5} + 5}{4} = \frac{3 + \sqrt{5}}{2} = 1 + φ

φ ϕ = \frac{1 + \sqrt{5}}{2} \frac{1 - \sqrt{5}}{2} = \frac{1 + 5}{4} = - 1

ϕ^{2} = {(\frac{1 - \sqrt{5}}{2})}^{2} = \frac{1 - 2 \sqrt{5} + 5}{4} = \frac{3 - \sqrt{5}}{2} = 1 + ϕ

φ + ϕ = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = 1

Nyní můžeme přejít k posloupnosti:

\begin{aligned} F_{0}^{2} + F_{1}^{2} + F_{2}^{2} + \dots + F_{n}^{2} \\ = \sum_{i = 0}^{n} F_{n}^{2} \\ = \sum_{i = 0}^{n} {(\frac{φ^{n} - ϕ^{n}}{\sqrt{5}})}^{2} \\ = \frac{1}{5} \sum_{i = 0}^{n} (φ^{2 n} - 2 φ^{n} ϕ^{n} + ϕ^{2 n}) \\ = \frac{1}{5} (\frac{1 - φ^{2 n + 2}}{1 - φ^{2}} - 2 \frac{1 - {(φ ϕ)}^{n + 1}}{1 - φ ϕ} + \frac{1 - ϕ^{2 n + 2}}{1 - ϕ^{2}}) \\ = \frac{1}{5} (\frac{1 - φ^{2 n + 2}}{- φ} - \frac{1 - (- 1)^{n + 1}}{1} + \frac{1 - ϕ^{2 n + 2}}{- ϕ}) \\ = \frac{φ^{2 n + 1} - (- 1)^{n} + ϕ^{2 n + 1}}{5} \\ = F_{n} F_{n + 1} \end{aligned}