Question

The value of $E=\frac{(1+17)(1+\frac{17}{2})(1+\frac{17}{3})......(1+\frac{17}{19})}{(1+19)(1+\frac{19}{2})(1+\frac{19}{3}).....(1+\frac{19}{17})}$ is,

• A. $1$
• B. ${}^{36}{C}_{17}$
• C. $\frac{2}{19}$
• D. ${}^{36}{C}_{18}$

Answer 1

Answer: (A)

$E=\frac{(1+17)(1+\frac{17}{2})(1+\frac{17}{3})......(1+\frac{17}{19})}{(1+19)(1+\frac{19}{2})(1+\frac{19}{3}).....(1+\frac{19}{17})}$

$=\frac{\frac{18\cdot 19\cdot \mathrm{20.............36}}{1\cdot 2\cdot 3\cdot \mathrm{4...............19}}}{\frac{20\cdot 21\cdot \mathrm{22.............36}}{1\cdot 2\cdot 3\cdot \mathrm{4...............17}}}$

$=\frac{18\cdot 19\cdot \mathrm{20.............36}}{1\cdot 2\cdot 3\cdot \mathrm{4...............19}}\times \frac{1\cdot 2\cdot 3\cdot \mathrm{4...............17}}{20\cdot 21\cdot \mathrm{22.............36}}$

$=\frac{18\cdot 19\cdot \mathrm{20.............36}}{20\cdot 21\cdot \mathrm{22.............36}}\times \frac{1\cdot 2\cdot 3\cdot \mathrm{4...............17}}{1\cdot 2\cdot 3\cdot \mathrm{4...............19}}$

$=\frac{18\cdot 19}{1}\times \frac{1}{18\cdot 19}=1$