## Calculus: Early Transcendentals (2nd Edition)

a) $243$ and $729$ b) $a_{n+1} = 3a_n$, $a_1 = 1$ c) $a_n = 3^{n-1}$
a) Another way of writing the sequence would be $\{3^0, 3^1, 3^2, 3^3, 3^4, ...\}$. The next two terms would be $3^5$ and $3^6$ or $243$ and $729$. . b) The first term is 1, so $a_1 = 1$ We have to multiply three to the previous term: $a_{n+1} = 3a_n$ c) We have to multiply $3$ by itself $n-1$ number of times. $a_n = 3^n$ would be incorrect because the first term ($a_1$) would be $3^1 = 3$ instead of $1$. $a_n = 3^{n-1}$