Exercise 7 Page 447

We are asked to write the first

6

a_{1} a_{n} = 2 = (a_{n - 1})^{2} + 1

To do so, we will use a table.

$n$	$a_{n} = (a_{n - 1})^{2} + 1$	$(a_{n - 1})^{2} + 1$	$a_{n}$
$1$	$a_{1} = 2$	$-$	$2$
$2$	$a_{2} = (a_{2 - 1})^{2} + 1$	$(a_{1})^{2} + 1 = 2^{2} + 1$	$5$
$3$	$a_{3} = (a_{3 - 1})^{2} + 1$	$(a_{2})^{2} + 1 = 5^{2} + 1$	$26$
$4$	$a_{4} = (a_{4 - 1})^{2} + 1$	$(a_{3})^{2} + 1 = 26^{2} + 1$	$677$
$5$	$a_{5} = (a_{5 - 1})^{2} + 1$	$(a_{4})^{2} + 1 = 677^{2} + 1$	$458330$
$6$	$a_{6} = (a_{6 - 1})^{2} + 1$	$(a_{5})^{2} + 1 = 458330^{2} + 1$	$210066388901$

Therefore, the first $6$ terms of the sequence are $2, 5, 26, 677, 458330$ and $210066388901 .$