The sequence is defined recursively. Compute the first five terms of the sequence. $a _ { 1 } = - 2 ; a _ { n } = \sqrt { a _ { 2 } ^ { n - 1 } + 1 } , n \geq 2$
A) $- 2 , \sqrt { 5 } , \sqrt { 6 } , \sqrt { 5 } , \sqrt { 2 }$
B) $- 2 , \sqrt { 10 } , \sqrt { 6 } , \sqrt { 7 } , 4 \sqrt { 2 }$
C) $- 2 , \sqrt { 5 } , \sqrt { 6 } , \sqrt { 7 } , 2 \sqrt { 2 }$
D) $- 2 , \sqrt { 5 } , \sqrt { 3 } , \sqrt { 7 } , 2 \sqrt { 8 }$
E) $- 2 , \sqrt { 5 } , \sqrt { 3 } , \sqrt { 7 } , \sqrt { 2 }$

Question

Quizplus · Accepted Answer

The sequence starts with $a_1 = -2$. For $n \geq 2$, the recursive formula is $a_n = \sqrt{a_{n-1}^2 + 1}$. Calculating the first five terms:- $a_1 = -2$- $a_2 = \sqrt{(-2)^2 + 1} = \sqrt{4 + 1} = \sqrt{5}$- $a_3 = \sqrt{(\sqrt{5})^2 + 1} = \sqrt{5 + 1} = \sqrt{6}$- $a_4 = \sqrt{(\sqrt{6})^2 + 1} = \sqrt{6 + 1} = \sqrt{7}$- $a_5 = \sqrt{(\sqrt{7})^2 + 1} = \sqrt{7 + 1} = \sqrt{8} = 2\sqrt{2}$

The Sequence Is Defined Recursively a1=−2;an=a2n−1+1,n≥2a _ { 1 } = - 2 ; a _ { n } = \sqrt { a _ { 2 } ^ { n - 1 } + 1 } , n \geq 2a1​=−2;an​=a2n−1​+1​,n≥2

The Sequence Is Defined Recursively $a _ { 1 } = - 2 ; a _ { n } = \sqrt { a _ { 2 } ^ { n - 1 } + 1 } , n \geq 2$