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# Solve the following question:When x=2011, the value of $\dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}}$ is ____.

Last updated date: 24th Jun 2024
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Hint: An expression is given and it has four divisions. It has a division within a division with a division within a division. So first solve the lowest division and after solving it go to the next one. Solve the equation by putting x as it is and then substitute the value of x.

We are given an expression $\dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}}$ . We have to find its value when the value of x is 2011
So the given expression has four divisions.
So the first division to be solved is $1 + \dfrac{1}{x}$
$1 + \dfrac{1}{x} = \dfrac{{x + 1}}{x}$
After substituting the division value, the given expression will become
$\Rightarrow \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}} = \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{\left( {\dfrac{{x + 1}}{x}} \right)}}}}}} = \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{x}{{x + 1}}}}}}$
The next division is $1 + \dfrac{x}{{x + 1}}$
$1 + \dfrac{x}{{x + 1}} = \dfrac{{\left( {x + 1} \right) + x}}{{x + 1}} = \dfrac{{2x + 1}}{{x + 1}}$
On substituting the above division value, we get
$\Rightarrow \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{x}{{x + 1}}}}}} = \dfrac{1}{{1 + \dfrac{1}{{\left( {\dfrac{{2x + 1}}{{x + 1}}} \right)}}}} = \dfrac{1}{{\left( {1 + \dfrac{{x + 1}}{{2x + 1}}} \right)}}$
The next division to be solved is $1 + \dfrac{{x + 1}}{{2x + 1}}$
$\Rightarrow 1 + \dfrac{{x + 1}}{{2x + 1}} = \dfrac{{\left( {2x + 1} \right) + x + 1}}{{2x + 1}} = \dfrac{{3x + 2}}{{2x + 1}}$
Substituting the above division value in the expression, we get
$\Rightarrow \dfrac{1}{{\left( {1 + \dfrac{{x + 1}}{{2x + 1}}} \right)}} = \dfrac{1}{{\left( {\dfrac{{3x + 2}}{{2x + 1}}} \right)}} = \dfrac{{2x + 1}}{{3x + 2}}$
Therefore, the value of the expression $\dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}}$ is $\dfrac{{2x + 1}}{{3x + 2}}$
We are asked to find its value when x is equal to 2011.
So we are substituting 2011 in the place of x.
$\Rightarrow \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{2011}}}}}}}} = \dfrac{{2\left( {2011} \right) + 1}}{{3\left( {2011} \right) + 2}} \\ = \dfrac{{4022 + 1}}{{6033 + 2}} \\ = \dfrac{{4023}}{{6035}} \\$
Therefore, the value of $\dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{{1 + \dfrac{1}{x}}}}}}}$ when x is equal to 2011 is $\dfrac{{4023}}{{6035}}$ .
So, the correct answer is “ $\dfrac{{4023}}{{6035}}$ ”.

Additional Information: If the denominator of a fraction is zero, then the value of that fraction is infinity.

Note: The reciprocal of a number is the fraction in which the numerator is 1 and the denominator is the number itself. The reciprocal of a reciprocal of a number is the number itself. That is what we have done in the above solution. Let a fraction be $\dfrac{a}{b}$ , then the reciprocal of this fraction will be $\dfrac{b}{a}$ . The numerator of the original fraction will become the denominator in its reciprocal and the denominator of the original fraction will become the numerator in its reciprocal.