Question
Convert the following products into factorials:
1.3.5.7.9....(2n - 1)

Answer

We have,
$1 \times 3 \times 5 \times 7 \times 9 \times.........(2\text{n} - 1)$
$=\frac{\big[1.3.5.7.9.....(2\text{n}-1)\big].\big[2.4.6.8.....(2\text{n}-2)(2\text{n})\big]}{2.4.6.8.....(2\text{n}-2)(2\text{n})}$
$=\frac{\big[1.3.5.7.9.....(2\text{n}-1)\big].\big[2.4.6.8.....(2\text{n}-2)(2\text{n})\big]}{2^\text{n}\big[1.2.3.4.......((\text{n-1})(\text{n}))\big]}$
$=\frac{1.2.3.4.5.6.7.8.......(2\text{n-2})(2\text{n-1})(2\text{n})}{2^\text{n}.\text{n}!}$
$=\frac{(2\text{n})!}{2^\text{n}.!}$
$\therefore 1.3.5.7.9......(2\text{n}-1)=\frac{(2\text{n})!}{2^\text{n}.\text{n!}}$

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