Question

Solve these both question of infinite series.

This is complex numbers question given in my book.

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Answer 1

Solution 1 :

Let's assume that,

$\rm x = 2 + \dfrac{1}{2 + \dfrac{1}{2 +... \infty } } ----(1)$

Now,

$\implies \rm x = 2 + \dfrac{1}{2 + \dfrac{1}{2 +... \infty } }$

${\implies \rm x = 2 + \dfrac{1}{x } \qquad(equation \: 1)}$

$\implies \rm x = \dfrac{2x + 1}{x }$

$\implies \rm x^{2} = 2x + 1$

$\implies \rm x^{2} - 2x - 1 = 0$

Now, solving it using quadratic formula.

$\implies \boxed{\rm x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}}$

$\rm \implies x = \dfrac{ - ( - 2) \pm \sqrt{ {( - 2)}^{2} - 4(1)( - 1) } }{2(1)}$

$\rm \implies x = \dfrac{ 2 \pm \sqrt{ 4 + 4} }{2(1)}$

$\rm \implies x = \dfrac{ 2 \pm \sqrt{ 8} }{2}$

$\rm \implies x = \dfrac{ 2 \pm 2\sqrt{2} }{2}$

$\rm \implies x = 1 \pm \sqrt{2}$

Now, since value of x cannot be negative or smaller than 2, x = 1 - √2 is not possible. So the required answer is x = 1 + √2.

Option (B) is correct.

$\rule{300}{1}$

Solution 2 :

We have,

${\rm x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty} } } ---(1)}$

${ \longrightarrow \rm x = \sqrt{6 + x } \qquad(equation \: 1.)}$

Squaring both sides

${ \longrightarrow \rm {x}^{2} =6 + x }$

${ \longrightarrow \rm {x}^{2} - x - 6 =0}$

${ \longrightarrow \rm {x}^{2} - 3x + 2x - 6 =0}$

${ \longrightarrow \rm x(x - 3) + 2(x - 3) =0}$

${ \longrightarrow \rm (x - 3)(x + 2) =0}$

${ \longrightarrow \rm (x - 3) = 0 \: or \: (x + 2) =0}$

${ \longrightarrow \rm \boxed{ \rm x = 3} \: or\: \boxed{ \rm x = - 2}}$

Since, the sum of possible values in square root cannot be negative, x = -2 is rejected.

x = 3 is the correct answer.

Option (c) is correct.