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$(2n + 1) (2n + 3) (2n + 5) ....... (4n - 1)$ is equal to :
$\frac{{(4n)\,\,!}}{{{2^n}\,.\,\,(2n)\,\,!\,\,(2n)\,\,!}}$
$\frac{{(4n)\,\,!\,\,\,n\,\,!}}{{{2^n}\,.\,\,(2n)\,\,!\,\,(2n)\,\,!}}$
$\frac{{(4n)\,\,!\,\,\,n\,\,!}}{{(2n)\,\,!\,\,(2n)\,\,!}}$
$\frac{{(4n)\,\,!\,\,\,n\,\,!}}{{{2^n}\,\,!\,\,(2n)\,\,!}}$
Solution
$E = (2 n + 1) (2 n + 3) (2 n + 5) ……(4 n – 1)$
Multiply numerator and denominator by $(2 n + 2) (2 n + 4) …… (4 n)$ & also by $(2 n ) !$ .
$E = \frac{{(2\,n)\,\,!\,\,\,(2\,n\,\, + \,\,1)\,\,(2\,n\,\, + \,\,2)\,\,(2\,n\,\, + \,\,3)\,\,……..\,\,(4\,n\, – \,1)\,\,.\,\,4\,n}}{{(2\,n)\,\,!\,\,\,(2\,n\, + \,2)\,\,(2\,n\, + \,4)\,\,……..\,\,(2\,n\, + \,2\,n)}}$
$= \frac{{(4\,n)\,\,!\,\,\,\,\,\,\,\,\,\,\,\, \times \,\,\,\,\,\,(n\,)\,\,!}}{{(2\,n)\,\,!\,\,\,{2^n}\,\,\left[ {(n\,\, + \,\,1)\,\,(n\,\, + \,\,2)\,\,……\,\,(2\,n)\,} \right]\,\,n\,\,!}}$ $= \frac{{(n\,\,!)\,\,.\,\,(4\,n)\,\,!}}{{{2^n}\,\,.\,\,{{\left( {(2\,n)\,\,!} \right)}^2}}}$
