the following sequence, defined by a recurrence relation, is monotonic and bounded, and therefore converges…

the following sequence, defined by a recurrence relation, is monotonic and bounded, and therefore converges. a. examine the first three terms of the sequence to determine whether the sequence is nondecreasing or nonincreasing. b. use analytical methods to find the limit of the sequence. a_n+1 = sqrt(35 + 2a_n); a_0 = -17 a. the first term is given. find the second and third terms of the sequence a_0 = -17 a_1 = a_2 = (type integers or decimals rounded to three decimal places as needed.)

the following sequence, defined by a recurrence relation, is monotonic and bounded, and therefore converges. a. examine the first three terms of the sequence to determine whether the sequence is nondecreasing or nonincreasing. b. use analytical methods to find the limit of the sequence. a_n+1 = sqrt(35 + 2a_n); a_0 = -17 a. the first term is given. find the second and third terms of the sequence a_0 = -17 a_1 = a_2 = (type integers or decimals rounded to three decimal places as needed.)

Answer

Explanation:

Step1: Calculate the second - term

Given the recurrence relation $a_{n + 1}=\sqrt{35+2a_{n}}$ and $a_0=-17$. Substitute $n = 0$ into the recurrence relation: $a_1=\sqrt{35+2a_0}=\sqrt{35+2\times(-17)}=\sqrt{35 - 34}=1$

Step2: Calculate the third - term

Substitute $n = 1$ into the recurrence relation. Since $a_1 = 1$, then $a_2=\sqrt{35+2a_1}=\sqrt{35+2\times1}=\sqrt{37}\approx6.083$

Answer:

$a_1 = 1$ $a_2\approx6.083$