Discrete Math On Recurrence

Question

1. 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.

Answer 1

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.