这是一份UCL伦敦大学 MATH00103作业代写的成功案例
translates into
$$
c \cdot q^{n+1}=c \cdot q^{n}+c \cdot q^{n-1}
$$
which after simplification becomes
$$
q^{2}=q+1
$$
So both numbers $c$ and $n$ disappear. ${ }^{1}$
So we have a quadratic equation for $q$, which we can solve and get
$$
q_{1}=\frac{1+\sqrt{5}}{2} \approx 1.618034, \quad q_{2}=\frac{1-\sqrt{5}}{2} \approx-0.618034 .
$$
This gives us two kinds of geometric progressions that satisfy the same recurrence as the Fibonacci numbers:
$$
G_{n}=c\left(\frac{1+\sqrt{5}}{2}\right)^{n}, \quad G_{n}^{\prime}=c\left(\frac{1-\sqrt{5}}{2}\right)^{n}
$$
MATH00103 COURSE NOTES ：
(a) if $a \mid b$ and $b \mid c$ then $a \mid c$;
(b) if $a \mid b$ and $a \mid c$ then $a \mid b+c$ and $a \mid b-c$;
(c) if $a, b>0$ and $a \mid b$ then $a \leq b$;
(d) if $a \mid b$ and $b \mid a$ then either $a=b$ or $a=-b$.