Question:
Prove that: $\frac{(2 n+1) !}{n !}=2^{n}[1 \cdot 3 \cdot 5 \ldots(2 n-1)(2 n+1)]$
Solution:
$\mathrm{LHS}=\frac{(2 n+1) !}{n !}$
$=\frac{(2 n+1)(2 n)(2 n-1) \ldots .(4)(3)(2)(1)}{n !}$
$=\frac{[(1)(3)(5) \ldots \ldots \ldots .(2 n-1)(2 n+1)][(2)(4)(6) \ldots \ldots \ldots(2 n)]}{n !}$
$=\frac{2^{n}[(1)(3)(5) \ldots \ldots \ldots(2 n-1)(2 n+1)][(1)(2)(3) \ldots \ldots \ldots(n)]}{n !}$
$=\frac{2^{n}[(1)(3)(5) \ldots \ldots \ldots(2 n-1)(2 n+1)][n !]}{n !}$
$=2^{n}[(1)(3)(5) \ldots \ldots \ldots(2 n-1)(2 n+1)]=\mathrm{RHS}$
Hence, proved.
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