The value of x + {x² + {x^4 + {x^8 + {x¹6} + }}}}

Correct Answer is : x(3+52)x≤ft({3 + {5}}{2})x(23+5)

Correct Answer is : x(3+52)x≤ft({3 + {5}}{2})x(23+5)

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SSC CGL Model Paper 3

Question : 64 of 100

Marks: +1, -0

The value of

x + \sqrt{x^2 + \sqrt{x^4 + \sqrt{x^8 + \sqrt{x^{16} + \dots}}}}

Solution:

x + \sqrt{x^2 + \sqrt{x^4 + \sqrt{x^8 + \sqrt{x^{16} + \dots}}}}

x + \sqrt{x^2 + \sqrt{x^4 + \sqrt{x^8 + x^8 \sqrt{1 + \dots}}}}

x + \sqrt{x^2 + \sqrt{x^4 + \sqrt{x^8(1 + \sqrt{ + \dots)}}}}

x + \sqrt{x^2 + \sqrt{x^4 + x^4 \sqrt{1 + \sqrt{1}}}}

x + \sqrt{x^2 + \sqrt{x^4(1 + \sqrt{1 + \dots)}}}

x + \sqrt{x^2 + x^2 \sqrt{1 + \sqrt{1 + \dots}}}

x + x \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}

x(1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots)}}}

Now

Let

1 + \sqrt{1 \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}}

= = y (let)

∴

1 + \sqrt{y} = y \Rightarrow \sqrt{y} = y - 1

Squaring both sides

y = y^2 + 1 - 2y

⇒

y^2 - 3y + 1 = 0

Then

y = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}

∵

1 + \sqrt{1 + \sqrt{1 + \dots}}

will be positive and equal to

\frac{3 + \sqrt{5}}{2}

∴

x\left(1 + \sqrt{1 + \sqrt{1 + \dots}}\right)

= x\left(\frac{3 + \sqrt{5}}{2}\right)

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