Answer: Option D
$$\left( {{\text{1 + }}\frac{1}{x}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)$$  $$\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)$$  $$\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right)$$
Taking L.C.M of each term
$$ \Rightarrow \left( {\frac{{x + 1}}{x}} \right)$$ $$\left( {\frac{{x + 1 + 1}}{{x + 1}}} \right)$$  $$\left( {\frac{{x + 2 + 1}}{{x + 2}}} \right)$$  $$\left( {\frac{{x + 3 + 1}}{{x + 3}}} \right)$$
$$\eqalign{
& \Rightarrow \frac{1}{x} \times \left( {x + 4} \right) \cr
& \Rightarrow \frac{{x + 4}}{x} \cr} $$