If pr-qr denote the rth convergent to -5-1-2- show that p3-p5-p2n-1-p2n-p2- q3-q5-q2n-1-q2n-q2-

We ca write √( 5 ) +1 2 =1+√( 5 ) 1 2 =1+ 2 √( 5 ) +1 ∴√( 5 ) +1 2 =1+ 1 1+ 1 1+ 1 1+ ...... Hence, p _ n+1 = p _ n + p _ n 1 p ...

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Quantitative Aptitude

Ratio & Proportion

If pr/qr de...

Question

If $\frac{p_{r}}{q_{r}}$ denote the $r^{t h}$ convergent to $\frac{\sqrt{5} + 1}{2},$ show that $p_{3} + p_{5} + . . . . + p_{2 n - 1} = p_{2 n} - p_{2}, q_{3} + q_{5} + . . . . + q_{2 n - 1} = q_{2 n} = q_{2}$ .

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Solution

We ca write $\frac{\sqrt{5} + 1}{2} = 1 + \frac{\sqrt{5} - 1}{2} = 1 + \frac{2}{\sqrt{5} + 1}$
$∴ \frac{\sqrt{5} + 1}{2} = 1 + \frac{1}{1 +} \frac{1}{1 +} \frac{1}{1 +} . . . . . .$
Hence, $p_{n + 1} = p_{n} + p_{n - 1}$
$p_{n + 1} - p_{n - 1} = p_{n}$
$p_{3} = p_{4} - p_{2}$
$p_{5} = p_{6} - p_{7}$
$p_{7} = p_{8} - p_{6}$
.
.
.
$p_{2 n - 1} = p_{2 n} - p_{2 n - 1}$
$∴ p_{3} + p_{5} + p_{7} + . . . . . . . + p_{2 n - 1} = p_{2 n} - p_{2}$
Similarly, $q_{n} = q_{n + 1} - q_{n - 1}$
Therefore, $q_{3} + q_{5} + q_{7} + . . . . . + q_{2 n - 1} = q_{2 n} - q_{2}$

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If prqr denote the rth convergent to √5+12, show that p3+p5+....+p2n−1=p2n−p2,q3+q5+....+q2n−1=q2n=q2.

We ca write √5+12=1+√5−12=1+2√5+1∴√5+12=1+11+11+11+......Hence, pn+1=pn+pn−1pn+1−pn−1=pnp3=p4−p2p5=p6−p7p7=p8−p6...p2n−1=p2n−p2n−1∴p3+p5+p7+.......+p2n−1=p2n−p2Similarly, qn=qn+1−qn−1Therefore, q3+q5+q7+.....+q2n−1=q2n−q2

If $\frac{p_{r}}{q_{r}}$ denote the $r^{t h}$ convergent to $\frac{\sqrt{5} + 1}{2},$ show that $p_{3} + p_{5} + . . . . + p_{2 n - 1} = p_{2 n} - p_{2}, q_{3} + q_{5} + . . . . + q_{2 n - 1} = q_{2 n} = q_{2}$ .