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The value of $$\left( {{\text{1 + }}\frac{1}{x}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right)$$ is?
Answer & Solution
Correct 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
$$ \Rightarrow \left( {\frac{{x + 1}}{x}} \right)$$ $$\left( {\frac{{x + 1 + 1}}{{x + 1}}} \right)$$ $$\left( {\frac{{x + 2 + 1}}{{x + 2}}} \right)$$ $$\left( {\frac{{x + 3 + 1}}{{x + 3}}} \right)$$
$$\eqalign{ & \Rightarrow \frac{1}{x} \times \left( {x + 4} \right) \cr & \Rightarrow \frac{{x + 4}}{x} \cr} $$
Taking L.C.M of each term
$$ \Rightarrow \left( {\frac{{x + 1}}{x}} \right)$$ $$\left( {\frac{{x + 1 + 1}}{{x + 1}}} \right)$$ $$\left( {\frac{{x + 2 + 1}}{{x + 2}}} \right)$$ $$\left( {\frac{{x + 3 + 1}}{{x + 3}}} \right)$$
$$\eqalign{ & \Rightarrow \frac{1}{x} \times \left( {x + 4} \right) \cr & \Rightarrow \frac{{x + 4}}{x} \cr} $$
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