## Calculus: Early Transcendentals (2nd Edition)

a) $\frac{1}{32}$ and $\frac{1}{64}$ b) $a_{n+1} = \frac{1}{2}a_n$, $a_1 = 1$ c) $a_n = 2^{1-n}$
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. $a_n = 2^{1-n}$