Question

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

Answer 1

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.

Complete step-by-step answer:
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.