(2n + 1) (2n + 3) (2n + 5) ....... (4n - 1) i | vedclass

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Question

$E = (2 n + 1) (2 n + 3) (2 n + 5) ......(4 n - 1)$

Multiply numerator and denominator by $(2 n + 2) (2 n + 4) ...... (4 n)$ & also by $(2 n )

Vedclass · Accepted Answer

$\frac{{(4n)\,\,!\,\,\,n\,\,!}}{{{2^n}\,.\,\,(2n)\,\,!\,\,(2n)\,\,!}}$

Vedclass · Answer

$E = (2 n + 1) (2 n + 3) (2 n + 5) ......(4 n - 1)$

Multiply numerator and denominator by $(2 n + 2) (2 n + 4) ...... (4 n)$ & also by $(2 n ) !$ .

$E = \frac{{(2\,n)\,\,!\,\,\,(2\,n\,\, + \,\,1)\,\,(2\,n\,\, + \,\,2)\,\,(2\,n\,\, + \,\,3)\,\,........\,\,(4\,n\, - \,1)\,\,.\,\,4\,n}}{{(2\,n)\,\,!\,\,\,(2\,n\, + \,2)\,\,(2\,n\, + \,4)\,\,........\,\,(2\,n\, + \,2\,n)}}$

$= \frac{{(4\,n)\,\,!\,\,\,\,\,\,\,\,\,\,\,\, 	imes \,\,\,\,\,\,(n\,)\,\,!}}{{(2\,n)\,\,!\,\,\,{2^n}\,\,\left[ {(n\,\, + \,\,1)\,\,(n\,\, + \,\,2)\,\,......\,\,(2\,n)\,} ight]\,\,n\,\,!}}$ $= \frac{{(n\,\,!)\,\,.\,\,(4\,n)\,\,!}}{{{2^n}\,\,.\,\,{{\left( {(2\,n)\,\,!} ight)}^2}}}$

$(2n + 1) (2n + 3) (2n + 5) ....... (4n - 1)$ is equal to :

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