Show that in the continued fraction b 1 a 1 - b 2 a 2 - b 3 a 3 - - p n - a n p n-1 - b n p n-2 - q n - a n q n-1 - b n q n-2-

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Arithmetic Progression

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Question

Show that in the continued fraction
$\frac{b_{1}}{a_{1}} - \frac{b_{2}}{a_{2}} - \frac{b_{3}}{a_{3}} - \dots$ ,
$p_{n} = a_{n} p_{n - 1} - b_{n} p_{n - 2}, q_{n} = a_{n} q_{n - 1} - b_{n} q_{n - 2}$ .

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\sqrt{a^{2} + 1}

2 (a^{2} + 1) q_{n} = p_{n - 1} + p_{n + 1}, 2 p_{n} = q_{n - 1} + q_{n + 1}

\frac{b_{1}}{a_{1} -} \frac{b_{2}}{a_{2} -} \frac{b_{3}}{a_{3} -} \dots

p_{n}

q_{n}

n

\frac{1}{a +} \frac{1}{b +} \frac{1}{a +} \frac{1}{b +} . . . .

p_{n + 2} - (a b + 2) p_{n} + p_{n - 2} = 0, q_{n + 2} - (a b + 2) q_{n} + q_{n - 2} = 0

\frac{p_{n}}{q_{n}}

n^{t h}

\sqrt{a^{2} + 1}

\frac{p_{2}^{2} + p_{3}^{2} + . . . . + p_{n + 1}^{2}}{q_{2}^{2} + q_{3}^{2} + . . . . + q_{n + 1}^{2}} = \frac{p_{n} +_{1} p_{n + 2} - p_{1} p_{2}}{q_{n} +_{1} q_{n + 2} - q_{1} q_{2}}

\frac{b}{a +} \frac{b}{a +} \frac{b}{a +} \dots,

p_{n + 1} = b q_{n}

b q_{n + 1} - a p_{n + 1} = b^{2} q_{n - 1}

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