The value of 3 + 1 4 + 1 3 + 1 4 + 1 3 + … ∞ is equal to

Let x = 3 + 1 4 + 1 3 + 1 4 + 1 3 + … ∞ So, x = 3 + 1 4 + 1 x = 3 + 1 4 x + 1 x ⇒ x - 3 = x ( 4 x + 1 ) ⇒ ( 4 x + 1 ) ( x - 3 ) = x ⇒ 4 x 2 - 12 x + x - 3 = x ⇒ 4 x 2 - 12 x - 3 = 0 x = 12 ± ( 12 ) 2 + 12 × 4 2 × 4 = 12 ± 12 ( 16 ) 8 = 12 ± 4 × 2 3 8 = 3 ± 2 3 2 x = 3 2 ± 3 = 1 . 5 ± 3 . But only positive value is accepted So, x = 1 . 5 + 3

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Mathematics - Quadratic Equation Question with Solution | TestHub

MathematicsQuadratic EquationTheory of equationsMedium2 minPYQ_2021

MathematicsMediumsingle choice

The value of $3 + \frac{1}{4 + \frac{1}{3 + \frac{1}{4 + \frac{1}{3 + \dots \infty}}}}$ is equal to

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Answer:

Solution:

Let $x = 3 + \frac{1}{4 + \frac{1}{3 + \frac{1}{4 + \frac{1}{3 + \dots \infty}}}}$

So, $x = 3 + \frac{1}{4 + \frac{1}{x}} = 3 + \frac{1}{\frac{4 x + 1}{x}}$

$\Rightarrow (x - 3) = \frac{x}{(4 x + 1)}$

$\Rightarrow (4 x + 1) (x - 3) = x$

$\Rightarrow 4 x^{2} - 12 x + x - 3 = x$

$\Rightarrow 4 x^{2} - 12 x - 3 = 0$

$x = \frac{12 \pm \sqrt{(12)^{2} + 12 \times 4}}{2 \times 4} = \frac{12 \pm \sqrt{12 (16)}}{8}$

$= \frac{12 \pm 4 \times 2 \sqrt{3}}{8} = \frac{3 \pm 2 \sqrt{3}}{2}$

$x = \frac{3}{2} \pm \sqrt{3} = 1.5 \pm \sqrt{3} .$

But only positive value is accepted

So, $x = 1.5 + \sqrt{3}$

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Theory of equations

⏱ 2mℹ️ Source: PYQ_2021

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