Assume that the sequence converges and find its limit.
-$$a _ { 1 } = - 3 , a _ { n + 1 } = \sqrt { 8 + 2 a _ { n } }$$
A) -3
B) -9
C) 4
D) 8

Question

Quizplus · Accepted Answer

First, we notice that the sequence is recursively defined. We can try to find the limit $L$ by assuming that $L$ exists and setting $a_{n+1}=L$. Solving for $L$ gives:
$$L=\sqrt{8+2L}$$
Squaring both sides gives:
$$L^2=8+2L$$
Rearranging gives:
$$L^2-2L-8=0$$
Solving for $L$ using the quadratic formula gives $L=4$ or $L=-2$. However, since $a_1=-3$ and $a_n$ is always non-negative, we can eliminate the negative solution and conclude that the limit of the sequence is $L=4$. Therefore, the answer is C.

Assume That the Sequence Converges and Find Its Limit a1=−3,an+1=8+2ana _ { 1 } = - 3 , a _ { n + 1 } = \sqrt { 8 + 2 a _ { n } }a1​=−3,an+1​=8+2an​​

Assume That the Sequence Converges and Find Its Limit $a _ { 1 } = - 3 , a _ { n + 1 } = \sqrt { 8 + 2 a _ { n } }$