Chapter 11 - Infinite Sequences and Series - 11.1 Exercises - Page 724: 4

{1, $\frac{9}{5}$, 3, $\frac{81}{17}$, $\frac{81}{11}$}

Plug in $n = 1$: $a_1$ = $\frac{3^1}{1 + 2^1}$ = $1$ Plug in $n = 2$: $a_2$ = $\frac{3^2}{1 + 2^2}$ = $\frac{9}{1 + 4}$ = $\frac{9}{5}$ And doing the same for $n = 3, 4, 5$ We obtain $a_3$ $=$ $3$ $a_4$ $=$ $\frac{81}{17}$ $a_5 =$ $\frac{81}{11}$ So the first five terms of $(a_n)$ are {1, $\frac{9}{5}$, 3, $\frac{81}{17}$, $\frac{81}{11}$}