## Calculus: Early Transcendentals (2nd Edition)

a) $32$ and $64$ b) $a_{n+1} = 2a_n$, $a_1 = 1$ c) $a_n = 2^{n-1}$
a) The terms are increasing in powers of two. Thus, the next two terms are $32$ and $64$. b) The first term is $1$, so $a_1 = 1$ Since we're simply multiplying two to the previous term in order to get the next term, the recursive function would be: $$a_{n+1} = 2a_n$$ c) We are simply multiplying the number $2$, $n-1$ amount of times (since $2^0$ would yield us $0$, the first term of the sequence). $$a_n = 2^{n-1}$$