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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 :

Similar Questions

If ${(1 + x - 2{x^2})^6} = 1 + {a_1}x + {a_2}{x^2} + .... + {a_{12}}{x^{12}}$, then the expression ${a_2} + {a_4} + {a_6} + .... + {a_{12}}$ has the value

$\sum\limits_{k = 0}^{10} {^{20}{C_k} = } $

In the expansion of ${(1 + x)^n}$ the sum of coefficients of odd powers of $x$ is

If $f(y) = 1 - (y - 1) + {(y - 1)^2} - {(y - 1)^{^3}} + ... - {(y - 1)^{17}},$ then the coefficient of $y^2$ in it is

The sum to $(n + 1)$ terms of the following series $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + $..... is