In a continued fraction 1 a- 1 b- 1 a- 1 b- -show that p n-2 -ab-2 p n - p n-2 -0- q n-2 -ab-2 q n - q n-2 -0

To show that p_n+2 (ab+2)p_n+p_n 2 #160;= 0 ,q_n+2 (ab+2)q_n+q_n 2 #160;= 0 Since, we have #160; p_2n+1 #160;= ap_2n +p_2n 1 , and p_2n ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Progressions

In a continue...

Question

In a continued fraction
$\frac{1}{a +} \frac{1}{b +} \frac{1}{a +} \frac{1}{b +} . . . .$ ,
show that
$p_{n + 2} - (a b + 2) p_{n} + p_{n - 2} = 0, q_{n + 2} - (a b + 2) q_{n} + q_{n - 2} = 0$

Open in App

Solution

Suggest Corrections

Similar questions

\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}

\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}

\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{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}}

(p + q)^{n} - (n - 1) p q (p + q)^{n - 2} + \frac{(n - 2) (n - 3)}{⌊ 2} p^{2} q^{2} (p - + q)^{n - 4} - . . . . . .

\frac{p^{n + 1} - q^{n + 1}}{p - q}

Join BYJU'S Learning Program