Question

What are the first two terms of the sequence defined recursively by the formula

Answer 1

Answer: C.  $\sqrt2,\ 6$

Step-by-step explanation:

Given formula $a_n=(a_{n-1})^2+4$

when$a_4=1604$

Substitute n=4 in th given formula , we get

$a_4=(a_{4-1})^2+4\\\Rightarrow1604=(a_3)^2+4\\\Rightarrow(a_3)^2=1604-4\\\Rightarrow(a_3)^2=1600\\\Rightarrow\ a_3=40$

Substitute n=3 in the given formula, we get

$a_3=(a_{3-1})^2+4\\\Rightarrow40=(a_2)^2+4\\\Rightarrow(a_2)^2=40-4\\\Rightarrow(a_2)^2=36\\\Rightarrow\ a_2=6$

Substitute n=2 in the given formula, we get

$a_2=(a_{2-1})^2+4\\\Rightarrow6=(a_1)^2+4\\\Rightarrow(a_1)^2=6-4\\\Rightarrow(a_1)^2=2\\\Rightarrow\ a_1=\sqrt{2}$

Thus, the first two terms of the sequence= $\sqrt{2}\ and \ 6$

Answer 2

A3^2+4=1604

a3=40

a2^2+4=40

a2=6

a1^2+4=6

a1=√2

So the first two terms are √2 and 6, answer C