Discrete Math On Recurrence 
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*Suppose that a geometric sequence starts with and satisfies the recurrence $a_n = ra_{n -1}$ for every positive integer $n$.


a) Show that $a_n = a_0rⁿ$.
b) Find the 100th number in the sequence 3,6,12,24,48, … . 
I know this is a another recurrence problem but not sure now to start with this one.

a) Note that $\dfrac{a_{n}}{a_{n-1}}=r$ for all $n$ integer, hence
$$\frac{a_n}{a_{n-1}}\cdot\frac{a_{n-1}}{a_{n-2}}\cdots\frac{a_1}{a_0}=r\cdot r\cdots r=r^n,$$
but $\displaystyle\frac{a_n}{a_{n-1}}\cdot\frac{a_{n-1}}{a_{n-2}}\cdots\frac{a_1}{a_0}=\frac{a_n}{a_0}$, thus 
$$\frac{a_n}{a_0}=r^n\Rightarrow a_n=a_0r^n.$$
b) In this sequence $a_0=3$ and the 100th term is equal $a_{99}$, here $r=2$, hence
$$a_{99}=a_0r=3\cdot 2^{99}$$.
Therefore, the 100th term is equal to $3\cdot 2^{99}$. If you want, you can compute this. 
