(2n + 1) (2n + 3) (2n + 5) ....... (4n

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7.Binomial Theorem

normal

$(2n + 1) (2n + 3) (2n + 5) ....... (4n - 1)$ 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 $(2 n ) !$ .

$E = \frac{{(2\,n)\,\,!\,\,\,(2\,n\,\, + \,\,1)\,\,(2\,n\,\, + \,\,2)\,\,(2\,n\,\, + \,\,3)\,\,……..\,\,(4\,n\, – \,1)\,\,.\,\,4\,n}}{{(2\,n)\,\,!\,\,\,(2\,n\, + \,2)\,\,(2\,n\, + \,4)\,\,……..\,\,(2\,n\, + \,2\,n)}}$

$= \frac{{(4\,n)\,\,!\,\,\,\,\,\,\,\,\,\,\,\, \times \,\,\,\,\,\,(n\,)\,\,!}}{{(2\,n)\,\,!\,\,\,{2^n}\,\,\left[ {(n\,\, + \,\,1)\,\,(n\,\, + \,\,2)\,\,……\,\,(2\,n)\,} \right]\,\,n\,\,!}}$ $= \frac{{(n\,\,!)\,\,.\,\,(4\,n)\,\,!}}{{{2^n}\,\,.\,\,{{\left( {(2\,n)\,\,!} \right)}^2}}}$

Standard 11

Mathematics

$(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) …… (1 + x + x^2 + …… + x^{100})$ when written in the ascending power of $x$ then the highest exponent of $x$ is ______ .

normal

If $\sum\limits_{K = 1}^{12} {12K{.^{12}}{C_K}{.^{11}}{C_{K – 1}}} $ is equal to $\frac{{12 \times 21 \times 19 \times 17 \times …….. \times 3}}{{11!}} \times {2^{12}} \times p$ then $p$ is

normal

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

Solution

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

Solution

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