Question

(2n + 1) (2n + 3) (2n + 5)....(4n1) is equal to

Answer 1

Answer:Let SA=(2n+1)(2n+3)(2n+5)⋅(4n−1)  

Let SB=(2n+1)(2n+2)(2n+3)(2n+4)(2n+5)(2n+6)…(4n−1)(4n)

⟹SB=(4n)!(2n)!(E01)

⟹SB=(2n+1)(2n+3)(2n+5)⋯(4n−1)=SA(2n+2)(2n+4)(2n+6)⋯(4n)

⟹SB=SA⋅2(n+1)⋅2(n+2)⋅2(n+3)⋯2(2n)

⟹SB=SA⋅2n⋅(n+1)(n+2)(n+3)⋯(2n)

⟹SB=SA⋅2n⋅(2n)!n!=(4n)!(2n)![ From (E01)]

⟹SA=[(4n)!(2n)!]÷[2n⋅(2n)!n!]=(4n)!⋅n!2n⋅(2n)!⋅(2n)!

⟹SA=(4n2n)⋅n!2n

Step-by-step explanation:

Answer 2

Answer:

$\star$ $AnswEr$ $\star$

Correct answer = [tex]\frac{(4n)!n!} {2^{n}  .(2n)!(2n)!}[/tex]

Plz mark as brainlest.