Answer
a. Verified (see below)
b. See proof below
Work Step by Step
$\{x_{n}\}=1,3,7,17,39,...$
$\{y_{n}\}=1,2,5,12,17,...$
$ a.\quad$
Verify $x_{1}^{2}-2y_{1}^{2}=-1,\qquad x_{2}^{2}-2y^{2}=+1$
$1^{2}-2(1^{2})=1-2=-1 $
$3^{2}-2(2^{2})=9-8=+1$
(verified)
Verify that $(a+2b)^{2}-2(a+b)^{2}=\pm 1,\ \quad $if $a^{2}-2b^{2}=\pm 1 $
$a^{2}+4ab+4b^{2}-2(a^{2}+2ab+b^{2})=$
$=a^{2}+4ab+4b^{2}-2a^{2}-4ab-2b^{2}$
$=-a^{2}+2b^{2}$
$=-(a^{2}-2b^{2})\qquad $
(by assumption, the parentheses are either 1 or -1)
$=-(\pm 1)$
$=\mp 1\qquad $
(which is either 1 or -1)
Hence, verified.
$ b.\quad$
$\begin{align*}
r_{n}^{2}-2&=[\displaystyle \frac{a+2b}{a+b}]^{2}-2 & & \\
r_{n}^{2}-2& =\displaystyle \frac{ (a+2b)^{2}-2(a+b)^{2}}{(a+b)^{2}} &\text{ ... use the result of a.} & \\
r_{n}^{2}-2&=\displaystyle \frac{ -(a^{2}-2b^{2})}{y_{n}^{2}} &\text{ ... use the result of a.} & \\
r_{n}^{2}&=2+\displaystyle \frac{ \pm 1}{y_{n}^{2}} & \\
r_{n}&=\sqrt{2+\frac{ \pm 1}{y_{n}^{2}}} & \end{align*}$
As n grows very large, it seems that $y_{n}$ also grows very large,
$\{y_{n}\}=1,2,5,12,17,...$
We want to prove that from the third term onward, $y_{n}\geq n$.
Proof by induction:
Let $n=3$. The third denominator is $5$, so $y_{n}\geq n$, for $n=3$
(base step of induction)
(Induction step)
Assume that the the nth denominator is $\geq n.$
Let $r_{n}=\displaystyle \frac{a}{b}, \quad (x_{n}=a,y_{n}=b\geq n $ and $a\geq b)$
Then ,
the (n+1)st denominator $y_{n}$ equals $(a+b)\geq(b+b)=2b$
$y_{n+1}\geq 2b \qquad $(b is the nth denominator...)
$y_{n+1}\geq 2n\qquad $(by the induction step hypothesis)
$y_{n+1}\geq n+1 \qquad $(2n $\geq$ n+1 for $n\geq
3)$
which proves the statement.
Thus, $\displaystyle \frac{ \pm 1}{y_{n}^{2}}\rightarrow 0$, when $ n\rightarrow\infty$.
And, we have that
$\displaystyle \lim_{n\rightarrow\infty}r_{n}=\sqrt{2}$
