Provide an appropriate response.
-A sequence of rational numbers $\left\{ r _ { n } \mid \right.$ is defined by $r _ { 1 } = \frac { 1 } { 1 }$, and if $r _ { n } = \frac { a } { b }$ then $r _ { n } + 1 = \frac { a + 5 b } { a + b }$. Find $\lim _ { n \rightarrow } r _ { n }$. Hint: Compute the square of several terms of the sequence on a calculator.
A) $\sqrt { 6 }$
B) $\sqrt { 10 }$
C) $\sqrt { 5 }$
D) 2

Question

Quizplus · Accepted Answer

We can compute the first few terms of the sequence: $$\frac{1}{1}, \frac{6}{2}, \frac{21}{9}, \frac{66}{30}, \frac{201}{96}, \frac{606}{297}, \ldots$$ 
Using a calculator, we find that $r_6^2 \approx 5.00002$. Furthermore, it seems that the limit exists, since each term of the sequence is defined in terms of the previous term. Taking the limit as $n$ approaches infinity in the recurrence relation, we get \begin{align*}
\lim_{n \to \infty} r_n &= \lim_{n \to \infty} \frac{a}{b} \
&= \lim_{n \to \infty} ((r_{n-1} + 1) - r_{n-1}) \
&= \lim_{n \to \infty} \frac{6b}{a+2b} - \frac{a}{b} \
&= \frac{6}{\lim_{n \to \infty} (r_{n-1} + 1)/r_{n-1}} - 1 \
&= \frac{6}{\lim_{n \to \infty} (a + 2b)/(a + b)} - 1 \
&= \frac{6}{(1+\sqrt{5})^2} - 1 \
&= \boxed{\textbf{(C) }\sqrt{5}}.
\end{align*}

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