DEAR EXPERTS PLEASE SOLVE THIS SUM

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \left(1-\frac{1}{\mathrm{n}}\right)\left(1-\frac{1}{\mathrm{n}+1}\right)\left(1-\frac{1}{\mathrm{n}+2}\right).....\left(1-\frac{1}{2\mathrm{n}}\right) \\ \left(1-\frac{1}{\mathrm{n}}\right)\left(1-\frac{1}{\mathrm{n}+1}\right)\left(1-\frac{1}{\mathrm{n}+2}\right).....\left(1-\frac{1}{2\mathrm{n}-1}\right)\left(1-\frac{1}{2\mathrm{n}}\right) \\ =\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)\left(\frac{\mathrm{n}+1-1}{\mathrm{n}+1}\right)\left(\frac{\mathrm{n}+2-1}{\mathrm{n}+2}\right).....\left(\frac{2\mathrm{n}-1-1}{2\mathrm{n}-1}\right)\left(\frac{2\mathrm{n}-1}{2\mathrm{n}}\right) \\ =\left(\frac{\mathrm{n}-1}{\cancel{\mathrm{n}}}\right)\left(\frac{\cancel{\mathrm{n}}}{\cancel{\mathrm{n}+1}}\right)\left(\frac{\cancel{\mathrm{n}+1}}{\cancel{\mathrm{n}+2}}\right)....\left(\frac{\cancel{2\mathrm{n}-2}}{\cancel{2\mathrm{n}-1}}\right)\left(\frac{\cancel{2\mathrm{n}-1}}{2\mathrm{n}}\right) \\ =\frac{\mathrm{n}-1}{2\mathrm{n}} \end{gathered}$$

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