Chapter 3 - Exercises - Page 140: 3b

Use the formula to find that $F(20)=6567$ --- Computing the same value of $F(20)$ just by plugging in 20 instead of $n$ in the formula: $F(n)=\frac{\sqrt{5}}{5} \cdot\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\frac{\sqrt{5}}{5} \cdot\left(\frac{1-\sqrt{5}}{2}\right)^{n}$ leads us to $F(20)=\frac{\sqrt{5}}{5} \cdot\left(\frac{1+\sqrt{5}}{2}\right)^{20}-\frac{\sqrt{5}}{5} \cdot\left(\frac{1-\sqrt{5}}{2}\right)^{20}=6567$

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