Question

$(1+1/x)(1+\frac1{{x+1}})(1+\frac1{x+2})(1+\frac1{x+3})=?$

• A. (x+4)/4
• B. (x+4)/x
• C. x
• D. 1

Answer 1

Answer:

B. (x+4)/x

$(1+1/x)(1+\frac1{{x+1}})(1+\frac1{x+2})(1+\frac1{x+3})$

$=\frac{x+1}{x}\times{\frac{x+1+1}{x+1}}\times{\frac{x+2+1}{x+2}}\times{\frac{x+3+1}{x+3}}$

$=\frac{x+1}{x}\times{\frac{x+2}{x+1}}\times{\frac{x+3}{x+2}}\times{\frac{x+4}{x+3}}$

$=\frac{x+4}{x}$