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 ...

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

Byju's Answer

Other

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 $p_{3} + p_{5} + . . . . + p_{2 n - 1} = p_{2 n} - p_{2}, q_{3} + q_{5} + . . . . + q_{2 n - 1} = q_{2 n} = q_{2}$ .

Open in App

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, $q_{3} + q_{5} + q_{7} + . . . . . + q_{2 n - 1} = q_{2 n} - q_{2}$

Suggest Corrections

Similar questions

Q. If

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

be the

n^{t h}

convergent to

\sqrt{a^{2} + 1}

, show that

\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}}

Q. If

\sqrt{N}

is converted into a continued fraction, and if n is the number of quotients in the period, show that

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

Q. If

P_{(2 n + 1, n - 1)} {: P}_{(2 n - 1, n)} = 3 : 5

then find

n

If the lines $p_{1} x + q_{1} y = 1, p_{2} x + q_{2} y = 1$ and $p_{3} x + q_{3} y = 1$ be concurrent, show that the points $(p_{1}, q_{1}), (p_{2}, q_{2})$ and $p_{3}, q_{3})$ are coolinear.

Q. If the lines p₁ x + q₁ y = 1, p₂ x + q₂ y = 1 and p₃ x + q₃ y = 1 be concurrent, show that the points (p₁, q₁), (p₂, q₂) and (p₃, q₃) are collinear.

Join BYJU'S Learning Program

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 $p_{3} + p_{5} + . . . . + p_{2 n - 1} = p_{2 n} - p_{2}, q_{3} + q_{5} + . . . . + q_{2 n - 1} = q_{2 n} = q_{2}$ .