G4.4 Given that $f_{1}=0, f_{2}=1$, and for any positive integer $n \geq 3, f_{n}=f_{n-1}+2 f_{n-2}$. If $d=f_{10}$, find the value of $d$.
The characteristic equation: $x^{2}=x+2 \Rightarrow x^{2}-x-2=0 \Rightarrow(x+1)(x-2)=0 \Rightarrow x=-1$ or 2
$$
\begin{array}{l}
f_{n}=A(-1)^{n}+B \times 2^{n}, n=1,2,3, \cdots \cdots \cdots \\
f_{1}=-A+2 B=0 \cdots \cdots \cdots(1) \\
f_{2}=A+4 B=1 \quad \cdots \cdots \cdots(2)
\end{array}
$$
(1) + (2) $6 B=1, B=\frac{1}{6}$

Sub. into (1): $-A+\frac{1}{3}=0, A=\frac{1}{3}$
$$
f_{n}=\frac{1}{3}(-1)^{n}+\frac{1}{6} \times 2^{n}, d=f_{10}=\frac{1}{3}+\frac{1}{6} \times 1024=\frac{513}{3}=171
$$

Method 2: $f_{1}=0, f_{2}=1$
$f_{3}=f_{2}+2 f_{1}=1+0=1 ; f_{4}=f_{3}+2 f_{2}=1+2=3$
$f_{5}=f_{4}+2 f_{3}=3+2 \times 1=5 ; f_{6}=f_{5}+2 f_{4}=5+2 \times 3=11$
$f_{7}=f_{6}+2 f_{5}=11+2 \times 5=21 ; f_{8}=f_{7}+2 f_{6}=21+2 \times 11=43$
$f_{9}=f_{8}+2 f_{7}=43+2 \times 21=85 ; f_{10}=f_{9}+2 f_{8}=85+2 \times 43=171=d$