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Prove that: `((2n)!)/(n !)={1. 3. 5 (2n-1)}2^ndot` |
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Answer» `L.H.S = ((2n)!)/(n!)` `= (1.2.3.4.5.6..(2n-1).(2n))/(n!)` `= ({2.4.6...(2n)}.{1.3.5....(2n-1)})/(n!)` `= (2^(n){1.2.3...n}{1.3.5...(2n-1)})/(n!)` `= 2^(n) {1.3.5..(2n-1)}` `= R.H.S` Hence Proved. |
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