Answer
a) $\frac{1}{32}$ and $\frac{1}{64}$
b) $a_{n+1} = \frac{1}{2}a_n $, $a_1 = 1$
c) $a_n = 2^{1-n}$
Work Step by Step
a) The denominator is increasing in powers of two. Thus, $16\times 2$ is the next term's denominator. $16\times2\times2$ is the following term's denominator. Thus, the next two terms are $\frac{1}{32}$ and $\frac{1}{64}$
b) The first term is $1$ so $a_1 = 1$.
The denominator is increasing in powers of two. We can alternatively say that the terms are increasing in powers of $\frac{1}{2}$.
For recursion, we just need to multiply $\frac{1}{2}$ to the previous term, and thus:
$a_{n+1} = \frac{1}{2}a_n $
c) Another way of writing out the sequence is $2^0, 2^{-1}, 2^{-2}, ...$
Since we start with $a_1$, we need to account for it. Thus, $a_n = a^{-n}$ is incorrect.
$a_n = 2^{1-n}$
