prove that: (2n+1) n! =2^n 1.3.5....(2n-1)(2n+1) - Vidyadip

Question

prove that: $\frac{(2\text{n+1})}{\text{n!}}=2^\text{n}\bigg\{1.3.5....(2\text{n}-1)(2\text{n}+1)\bigg\}$

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Answer

We have, $\text{L.H.S.}= \frac{(2\text{n+1})}{\text{n!}}$ $=\frac{(2\text{n+1)}\big[1.2.3.4.5.6.7.8.9........(2\text{n-1})\text{2n}\big]}{\text{n!}}$ $=\frac{\big[1.3.5.7....(2\text{n}-1)\times(2\text{n}+1)\big]\big[2.4.6.8.....(2\text{n-2})2\text{n}\big]}{\text{n!}}$ $=\frac{\big[1.3.5.7....(2\text{n-1})(2\text{n}+1)\big]\times 2^\text{n}\big[1.2.3.4.....(\text{n}-1) \text{n}\big]}{\text{n!}}$ $=\frac{\big[1.3.5.7.....(2\text{n}-1)(2\text{n}+1)\big]2^\text{n}\times\text{n!}}{\text{n!}}$ $= 2^\text{n} [ 1.3.5.7.......(2\text{n} - 1)(2\text{n} + 1)]$ $=\text{R.H.S.}$ Hence proved.

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