Let E -2 n -12 n -32 n -5 -4 n -34 n -1 where n -1- then 2 n E is divisible byA- 2 n C nB- 4 n C 2 nC- 4 n C -2D- 2 n C -2

The correct option is B ^4nC_2n E = (2n +1) (2n + 3) (2n + 5)⋯ (4n 3) (4n 1) E = (2n)!(2n+1)(2n+2)(2n+3)(2n+4)(2n+5)⋯(4n 1)4n /(2n)!(2n+2)(2n+4)⋯ (4n) E= ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

Let E =2 n +1...

Question

Let $E = (2 n + 1) (2 n + 3) (2 n + 5) \dots (4 n - 3) (4 n - 1)$ where $n > 1$ , then $2^{n}$ E is divisible by

$^{2 n} C_{n}$

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$^{4 n} C_{2 n}$

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$^{2 n} C_{\frac{n}{2}}$

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$^{4 n} C_{\frac{n}{2}}$

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Solution

The correct option is B
$^{4 n} C_{2 n}$

$E = (2 n + 1) (2 n + 3) (2 n + 5) \dots (4 n - 3) (4 n - 1) E = \frac{(2 n)! (2 n + 1) (2 n + 2) (2 n + 3) (2 n + 4) (2 n + 5) \dots (4 n - 1) 4 n}{(2 n)! (2 n + 2) (2 n + 4) \dots (4 n)} E = \frac{(4 n)! n!}{(2 n)! n! 2^{n} (n + 1) (n + 2) \dots (2 n)} E = \frac{(4 n)! n!}{(2 n)! (2 n)! 2^{n}} \Rightarrow 2^{n} E = n!^{4 n} C_{2 n}$
Hence $2^{n} E$ is divisible by $^{4 n} C_{2 n}$

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Let E=(2n+1)(2n+3)(2n+5)⋯(4n–3)(4n–1) where n>1, then 2n E is divisible by

The correct option is B 4nC2n E=(2n+1)(2n+3)(2n+5)⋯(4n−3)(4n−1)E=(2n)!(2n+1)(2n+2)(2n+3)(2n+4)(2n+5)⋯(4n−1)4n(2n)!(2n+2)(2n+4)⋯(4n)E=(4n)!n!(2n)!n!2n(n+1)(n+2)⋯(2n)E=(4n)!n!(2n)!(2n)!2n⇒2nE=n!4nC2n Hence 2nE is divisible by 4nC2n

Let $E = (2 n + 1) (2 n + 3) (2 n + 5) \dots (4 n - 3) (4 n - 1)$ where $n > 1$ , then $2^{n}$ E is divisible by