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 = 1 a_2 = 6.083 (type integers or decimals rounded to three decimal places as needed.) based on the first three terms of the sequence (a_0, a_1, and a_2) the sequence is monotonically nondecreasing. b. the limit of the sequence is (simplify your answer. type an integer or a simplified fraction.)

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 = 1 a_2 = 6.083 (type integers or decimals rounded to three decimal places as needed.) based on the first three terms of the sequence (a_0, a_1, and a_2) the sequence is monotonically nondecreasing. b. the limit of the sequence is (simplify your answer. type an integer or a simplified fraction.)

Answer

Explanation:

Step1: Assume the limit exists

Let $\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}a_{n + 1}=L$.

Step2: Substitute into the recurrence relation

Since $a_{n+1}=\sqrt{35 + 2a_{n}}$, we have $L=\sqrt{35+2L}$.

Step3: Square both sides

Squaring gives $L^{2}=35 + 2L$, which can be rewritten as $L^{2}-2L - 35=0$.

Step4: Solve the quadratic - equation

Using the quadratic formula $L=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-2$, $c=-35$), we get $L=\frac{2\pm\sqrt{(-2)^{2}-4\times1\times(-35)}}{2\times1}=\frac{2\pm\sqrt{4 + 140}}{2}=\frac{2\pm\sqrt{144}}{2}=\frac{2\pm12}{2}$.

Step5: Reject the non - valid solution

We have two solutions $L_1=\frac{2 + 12}{2}=7$ and $L_2=\frac{2-12}{2}=-5$. But since $a_{n+1}=\sqrt{35 + 2a_{n}}\geq0$ for all $n$ (the square - root function is non - negative), we reject $L=-5$.

Answer:

$7$