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?
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
Answer: Option D
Solution(By Examveda Team)
$$\left( {{\text{1 + }}\frac{1}{x}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right)$$Taking L.C.M of each term
$$ \Rightarrow \left( {\frac{{x + 1}}{x}} \right)$$ $$\left( {\frac{{x + 1 + 1}}{{x + 1}}} \right)$$ $$\left( {\frac{{x + 2 + 1}}{{x + 2}}} \right)$$ $$\left( {\frac{{x + 3 + 1}}{{x + 3}}} \right)$$
$$\eqalign{ & \Rightarrow \frac{1}{x} \times \left( {x + 4} \right) \cr & \Rightarrow \frac{{x + 4}}{x} \cr} $$
This Question Belongs to Arithmetic Ability >> Algebra
Join The Discussion
Comments ( 3 )
Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
in the third step the x in the numerators and denominators will cancel out each other so it will be 1*2*3/2*4/3=4//
What happened 3rd step
How this answer come we cannot understand it easily explain it more