A mind blowing series

Here $a$ and $b$ are constants where $a=3$ and $b=4$ .

A series is taken with first term as $\dfrac{a}{b}$ where $a$ and $b$ are coprime.

The second term is $\dfrac{a+2b}{a+3b}$ . This is taken equal to $\dfrac{c}{d}$ .

The next term is given by $\dfrac{c+2d}{c+3d}$ . This is again taken equal to $\dfrac{e}{f}$ .

The next term is given by $\dfrac{e+2f}{e+3f}$ .

This process is repeated $n$ times.

Can we find the $n^{th}$ term in terms of $a,b$ and $n$ ?

The series is $\dfrac{3}{4}, \dfrac{11}{15}, \dfrac{41}{56} \cdots \cdots\cdots \cdots \cdots\cdots\underbrace{n}_{\text{ times}}$

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

If the two roots of the equation $x^2-4x+1=0$ are $\alpha,\beta$ .

Define $c_n=\dfrac{\alpha^n-\beta^n}{\alpha-\beta}$ . ( $c_1=1,c_2=4,c_n=4c_{n-1}-c_{n-2}$ )

Then the $n^{th}$ term of the sequence is $\dfrac{(c_{n-1}-c_{n-2})a+2c_{n-1}b}{c_{n-1}a+(c_{n}-c_{n-1})b}$

X X - 3 years ago

Can you please provide a proof ?

Vinayak Bansal - 3 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$